loukas grafakos classical fourier analysis

Graduate Texts in Mathematics

Classical Fourier Analysis

Loukas Grafakos

Third Edition

Graduate Texts in Mathematics 249

Graduate Texts in Mathematics

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Loukas Grafakos

Classical Fourier Analysis

Third Edition

123

Loukas GrafakosDepartment of MathematicsUniversity of MissouriColumbia, MO, USA

ISSN 0072-5285 ISSN 2197-5612 (electronic)ISBN 978-1-4939-1193-6 ISBN 978-1-4939-1194-3 (eBook)DOI 10.1007/978-1-4939-1194-3Springer New York Heidelberg Dordrecht London

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To Suzanne

Preface

The great response to the publication of my book Classical and Modern Fourier

Analysis in 2004 has been especially gratifying to me. I was delighted when Springer

offered to publish the second edition in 2008 in two volumes: Classical Fourier

Analysis, 2nd Edition, and Modern Fourier Analysis, 2nd Edition. I am now elated

to have the opportunity to write the present third edition of these books, which

Springer has also kindly offered to publish. The third edition was born from my

desire to improve the exposition in several places, fix a few inaccuracies, and add

some new material. I have been very fortunate to receive several hundred e-mail

messages that helped me improve the proofs and locate mistakes and misprints in

the previous editions.

In this edition, I maintain the same style as in the previous ones. The proofs con-

tain details that unavoidably make the reading more cumbersome. Although it will

behoove many readers to skim through the more technical aspects of the presenta-

tion and concentrate on the flow of ideas, the fact that details are present will be

comforting to some. (This last sentence is based on my experience as a graduate

student.) Readers familiar with the second edition will notice that the chapter on

weights has been moved from the second volume to the first.

This first volume Classical Fourier Analysis is intended to serve as a text for

a one-semester course with prerequisites of measure theory, Lebesgue integration,

and complex variables. I am aware that this book contains significantly more ma-

terial than can be taught in a semester course; however, I hope that this additional

information will be useful to researchers. Based on my experience, the following list

of sections (or parts of them) could be taught in a semester without affecting the

logical coherence of the book: Sections 1.1, 1.2, 1.3, 2.1, 2.2., 2.3, 3.1, 3.2, 3.3, 4.4,

4.5, 5.1, 5.2, 5.3, 5.5, 5.6, 6.1, 6.2.

A long list of people have assisted me in the preparation of this book, but I remain

solely responsible for any misprints, mistakes, and omissions contained therein.

Please contact me directly ([email protected]) if you have corrections or com-

ments. Any corrections to this edition will be posted to the website

http://math.missouri.edu/˜loukas/FourierAnalysis.html

vii

http://math.missouri.edu/~loukas/FourierAnalysis.html

viii Preface

which I plan to update regularly. I have prepared solutions to all of the exercises for

the present edition which will be available to instructors who teach a course out of

this book.

Athens, Greece, Loukas Grafakos

March 2014

Acknowledgments

I am extremely fortunate that several people have pointed out errors, misprints, and

omissions in the previous editions of the books in this series. All these individuals

have provided me with invaluable help that resulted in the improved exposition of

the text. For these reasons, I would like to express my deep appreciation and sincere

gratitude to the all of the following people.

ix

First edition acknowledgements: Georgios Alexopoulos, Nakhle Asmar, Bruno

Calado, Carmen Chicone, David Cramer, Geoffrey Diestel, Jakub Duda, Brenda

Frazier, Derrick Hart, Mark Hoffmann, Steven Hofmann, Helge Holden, Brian

Hollenbeck, Petr Honzık, Alexander Iosevich, Tunde Jakab, Svante Janson, Ana

Jimenez del Toro, Gregory Jones, Nigel Kalton, Emmanouil Katsoprinakis, Dennis

Kletzing, Steven Krantz, Douglas Kurtz, George Lobell, Xiaochun Li, Jose Marıa

Martell, Antonios Melas, Keith Mersman, Stephen Montgomery-Smith, Andrea

Nahmod, Nguyen Cong Phuc, Krzysztof Oleszkiewicz, Cristina Pereyra, Carlos

Perez, Daniel Redmond, Jorge Rivera-Noriega, Dmitriy Ryabogin, Christopher

Sansing, Lynn Savino Wendel, Shih-Chi Shen, Roman Shvidkoy, Elias M. Stein,

Atanas Stefanov, Terence Tao, Erin Terwilleger, Christoph Thiele, Rodolfo Torres,

Deanie Tourville, Nikolaos Tzirakis, Don Vaught, Igor Verbitsky, Brett Wick, James

Wright, and Linqiao Zhao.

Second edition acknowledgements: Marco Annoni, Pascal Auscher, Andrew

Bailey, Dmitriy Bilyk, Marcin Bownik, Juan Cavero de Carondelet Fiscowich,

Leonardo Colzani, Simon Cowell, Mita Das, Geoffrey Diestel, Yong Ding, Jacek

Dziubanski, Frank Ganz, Frank McGuckin, Wei He, Petr Honzık, Heidi Hulsizer,

Philippe Jaming, Svante Janson, Ana Jimenez del Toro, John Kahl, Cornelia Kaiser,

Nigel Kalton, Kim Jin Myong, Doowon Koh, Elena Koutcherik, David Kramer,

Enrico Laeng, Sungyun Lee, Qifan Li, Chin-Cheng Lin, Liguang Liu, Stig-Olof

Londen, Diego Maldonado, Jose Marıa Martell, Mieczysław Mastyło, Parasar

Mohanty, Carlo Morpurgo, Andrew Morris, Mihail Mourgoglou, Virginia Naibo,

Tadahiro Oh, Marco Peloso, Maria Cristina Pereyra, Carlos Perez, Humberto Rafeiro,

Maria Carmen Reguera Rodrıguez, Alexander Samborskiy, Andreas Seeger, Steven

Senger, Sumi Seo, Christopher Shane, Shu Shen, Yoshihiro Sawano, Mark Spencer,

Vladimir Stepanov, Erin Terwilleger, Rodolfo H. Torres, Suzanne Tourville,

x Acknowledgments

Among all these people, I would like to give special thanks to an individual who

has studied extensively the two books in the series and has helped me more than

anyone else in the preparation of the third edition: Danqing He. I am indebted to him

for all the valuable corrections, suggestions, and constructive help he has provided

me with in this work. Without him, these books would have been a lot poorer.

Finally, I would also like to thank the University of Missouri for granting me

a research leave during the academic year 2013-2014. This time off enabled me to

finish the third edition of this book on time. I spent my leave in Greece.

Ignacio Uriarte-Tuero, Kunyang Wang, Huoxiong Wu, Kozo Yabuta, Takashi

Yamamoto, and Dachun Yang.

Third edition acknowledgments: Marco Annoni, Mark Ashbaugh, Daniel

Azagra, Andrew Bailey, Arpad Benyi, Dmitriy Bilyk, Nicholas Boros, Almut

Burchard, Marıa Carro, Jameson Cahill, Juan Cavero de Carondelet Fiscowich,

Xuemei Chen, Andrea Fraser, Shai Dekel, Fausto Di Biase, Zeev Ditzian, Jianfeng

Dong, Oliver Dragicevic, Sivaji Ganesh, Friedrich Gesztesy, Zhenyu Guo, Piotr

Hajłasz, Danqing He, Andreas Heinecke, Steven Hofmann, Takahisa Inui, Junxiong

Jia, Kasinathan Kamalavasanthi, Hans Koelsch, Richard Laugesen, Kaitlin Leach,

Andrei Lerner, Yiyu Liang, Calvin Lin, Liguang Liu, Elizabeth Loew, Chao Lu,

Richard Lynch, Diego Maldonado, Lech Maligranda, Richard Marcum, Mieczysław

Mastyło, Mariusz Mirek, Carlo Morpurgo, Virginia Naibo, Hanh Van Nguyen,

Seungly Oh, Tadahiro Oh, Yusuke Oi, Lucas da Silva Oliveira, Kevin O’Neil, Hesam

Oveys, Manos Papadakis, Marco Peloso, Carlos Perez, Jesse Peterson, Dmitry

Prokhorov, Amina Ravi, Maria Carmen Reguera Rodrıguez, Yoshihiro Sawano,

Mirye Shin, Javier Soria, Patrick Spencer, Marc Strauss, Krystal Taylor, Naohito

Tomita, Suzanne Tourville, Rodolfo H. Torres, Fujioka Tsubasa, Ignacio Uriarte-

Tuero, Brian Tuomanen, Shibi Vasudevan, Michael Wilson, Dachun Yang, Kai Yang,

Yandan Zhang, Fayou Zhao, and Lifeng Zhao.

Contents

1 Lp Spaces and Interpolation 1

1.1 Lp and Weak Lp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 The Distribution Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.2 Convergence in Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.1.3 A First Glimpse at Interpolation . . . . . . . . . . . . . . . . . . . . . . . . 9

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2 Convolution and Approximate Identities . . . . . . . . . . . . . . . . . . . . . . . 17

1.2.1 Examples of Topological Groups . . . . . . . . . . . . . . . . . . . . . . . 18

1.2.2 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.2.3 Basic Convolution Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.2.4 Approximate Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.3 Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.3.1 Real Method: The Marcinkiewicz Interpolation Theorem . . . 33

1.3.2 Complex Method: The Riesz–Thorin Interpolation

Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

1.3.3 Interpolation of Analytic Families of Operators . . . . . . . . . . . 40

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

1.4 Lorentz Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

1.4.1 Decreasing Rearrangements . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

1.4.2 Lorentz Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

1.4.3 Duals of Lorentz Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

1.4.4 The Off-Diagonal Marcinkiewicz Interpolation Theorem . . . 60

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

2 Maximal Functions, Fourier Transform, and Distributions 85

2.1 Maximal Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

2.1.1 The Hardy–Littlewood Maximal Operator . . . . . . . . . . . . . . . . 86

2.1.2 Control of Other Maximal Operators . . . . . . . . . . . . . . . . . . . . 90

xi

xii Contents

2.1.3 Applications to Differentiation Theory . . . . . . . . . . . . . . . . . . 93

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

2.2 The Schwartz Class and the Fourier Transform . . . . . . . . . . . . . . . . . . 104

2.2.1 The Class of Schwartz Functions . . . . . . . . . . . . . . . . . . . . . . . 105

2.2.2 The Fourier Transform of a Schwartz Function . . . . . . . . . . . 108

2.2.3 The Inverse Fourier Transform and Fourier Inversion . . . . . . 111

2.2.4 The Fourier Transform on L1 +L2 . . . . . . . . . . . . . . . . . . . . . . 113

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

2.3 The Class of Tempered Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . 119

2.3.1 Spaces of Test Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

2.3.2 Spaces of Functionals on Test Functions . . . . . . . . . . . . . . . . . 120

2.3.3 The Space of Tempered Distributions . . . . . . . . . . . . . . . . . . . 123

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

2.4 More About Distributions and the Fourier Transform . . . . . . . . . . . . . 133

2.4.1 Distributions Supported at a Point . . . . . . . . . . . . . . . . . . . . . . 134

2.4.2 The Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

2.4.3 Homogeneous Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

2.5 Convolution Operators on Lp Spaces and Multipliers . . . . . . . . . . . . . 146

2.5.1 Operators That Commute with Translations . . . . . . . . . . . . . . 146

2.5.2 The Transpose and the Adjoint of a Linear Operator . . . . . . . 150

2.5.3 The Spaces M p,q(Rn) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

2.5.4 Characterizations of M 1,1(Rn) and M 2,2(Rn) . . . . . . . . . . . . 153

2.5.5 The Space of Fourier Multipliers Mp(Rn) . . . . . . . . . . . . . . . 155

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

2.6 Oscillatory Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

2.6.1 Phases with No Critical Points . . . . . . . . . . . . . . . . . . . . . . . . . 161

2.6.2 Sublevel Set Estimates and the Van der Corput Lemma. . . . . 164

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

3 Fourier Series 173

3.1 Fourier Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

3.1.1 The n-Torus Tn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

3.1.2 Fourier Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

3.1.3 The Dirichlet and Fejer Kernels . . . . . . . . . . . . . . . . . . . . . . . . 178

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

3.2 Reproduction of Functions from Their Fourier Coefficients . . . . . . . . 183

3.2.1 Partial sums and Fourier inversion . . . . . . . . . . . . . . . . . . . . . . 183

3.2.2 Fourier series of square summable functions . . . . . . . . . . . . . 185

3.2.3 The Poisson Summation Formula . . . . . . . . . . . . . . . . . . . . . . . 187

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

3.3 Decay of Fourier Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

3.3.1 Decay of Fourier Coefficients of Arbitrary Integrable

Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

3.3.2 Decay of Fourier Coefficients of Smooth Functions . . . . . . . . 195

Contents xiii

3.3.3 Functions with Absolutely Summable Fourier

Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

3.4 Pointwise Convergence of Fourier Series . . . . . . . . . . . . . . . . . . . . . . . 204

3.4.1 Pointwise Convergence of the Fejer Means . . . . . . . . . . . . . . . 204

3.4.2 Almost Everywhere Convergence of the Fejer Means . . . . . . 207

3.4.3 Pointwise Divergence of the Dirichlet Means . . . . . . . . . . . . . 210

3.4.4 Pointwise Convergence of the Dirichlet Means . . . . . . . . . . . . 212

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

3.5 A Tauberian theorem and Functions of Bounded Variation . . . . . . . . 216

3.5.1 A Tauberian theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216

3.5.2 The sine integral function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

3.5.3 Further properties of functions of bounded variation . . . . . . . 219

3.5.4 Gibbs phenomenon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225

3.6 Lacunary Series and Sidon Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

3.6.1 Definition and Basic Properties of Lacunary Series . . . . . . . . 227

3.6.2 Equivalence of Lp Norms of Lacunary Series . . . . . . . . . . . . . 229

3.6.3 Sidon sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

4 Topics on Fourier Series 241

4.1 Convergence in Norm, Conjugate Function,

and Bochner–Riesz Means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

4.1.1 Equivalent Formulations of Convergence in Norm . . . . . . . . . 242

4.1.2 The Lp Boundedness of the Conjugate Function . . . . . . . . . . . 246

4.1.3 Bochner–Riesz Summability . . . . . . . . . . . . . . . . . . . . . . . . . . . 250

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

4.2 A. E. Divergence of Fourier Series and Bochner–Riesz means . . . . . 255

4.2.1 Divergence of Fourier Series of Integrable Functions . . . . . . 255

4.2.2 Divergence of Bochner–Riesz Means of Integrable

Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270

4.3 Multipliers, Transference, and Almost Everywhere Convergence . . . 271

4.3.1 Multipliers on the Torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

4.3.2 Transference of Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275

4.3.3 Applications of Transference . . . . . . . . . . . . . . . . . . . . . . . . . . . 280

4.3.4 Transference of Maximal Multipliers . . . . . . . . . . . . . . . . . . . . 281

4.3.5 Applications to Almost Everywhere Convergence . . . . . . . . . 285

4.3.6 Almost Everywhere Convergence of Square Dirichlet

Means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289

xiv Contents

4.4 Applications to Geometry and Partial Differential Equations . . . . . . . 292

4.4.1 The Isoperimetric Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . 292

4.4.2 The Heat Equation with Periodic Boundary Condition . . . . . 294

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298

4.5 Applications to Number theory and Ergodic theory . . . . . . . . . . . . . . 299

4.5.1 Evaluation of the Riemann Zeta Function at even Natural

numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

4.5.2 Equidistributed sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302

4.5.3 The Number of Lattice Points inside a Ball . . . . . . . . . . . . . . . 305

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308

5 Singular Integrals of Convolution Type 313

5.1 The Hilbert Transform and the Riesz Transforms . . . . . . . . . . . . . . . . 313

5.1.1 Definition and Basic Properties of the Hilbert Transform . . . 314

5.1.2 Connections with Analytic Functions . . . . . . . . . . . . . . . . . . . . 317

5.1.3 Lp Boundedness of the Hilbert Transform . . . . . . . . . . . . . . . . 319

5.1.4 The Riesz Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329

5.2 Homogeneous Singular Integrals and the Method of Rotations . . . . . 333

5.2.1 Homogeneous Singular and Maximal Singular Integrals . . . . 333

5.2.2 L2 Boundedness of Homogeneous Singular Integrals . . . . . . 336

5.2.3 The Method of Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339

5.2.4 Singular Integrals with Even Kernels . . . . . . . . . . . . . . . . . . . . 341

5.2.5 Maximal Singular Integrals with Even Kernels . . . . . . . . . . . 347

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353

5.3 The Calderon–Zygmund Decomposition and Singular Integrals . . . . 355

5.3.1 The Calderon–Zygmund Decomposition . . . . . . . . . . . . . . . . . 355

5.3.2 General Singular Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358

5.3.3 Lr Boundedness Implies Weak Type (1,1) Boundedness . . . . 359

5.3.4 Discussion on Maximal Singular Integrals . . . . . . . . . . . . . . . 362

5.3.5 Boundedness for Maximal Singular Integrals Implies

Weak Type (1,1) Boundedness . . . . . . . . . . . . . . . . . . . . . . . . . 366

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371

5.4 Sufficient Conditions for Lp Boundedness . . . . . . . . . . . . . . . . . . . . . . 374

5.4.1 Sufficient Conditions for Lp Boundedness of Singular

Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375

5.4.2 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378

5.4.3 Necessity of the Cancellation Condition . . . . . . . . . . . . . . . . . 379

5.4.4 Sufficient Conditions for Lp Boundedness of Maximal

Singular Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384

5.5 Vector-Valued Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385

5.5.1 ℓ2-Valued Extensions of Linear Operators . . . . . . . . . . . . . . . . 386

5.5.2 Applications and ℓr-Valued Extensions of Linear

Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390

Contents xv

5.5.3 General Banach-Valued Extensions . . . . . . . . . . . . . . . . . . . . . 391

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398

5.6 Vector-Valued Singular Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401

5.6.1 Banach-Valued Singular Integral Operators . . . . . . . . . . . . . . . 402

5.6.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408

5.6.3 Vector-Valued Estimates for Maximal Functions . . . . . . . . . . 411

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414

6 Littlewood–Paley Theory and Multipliers 419

6.1 Littlewood–Paley Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419

6.1.1 The Littlewood–Paley Theorem . . . . . . . . . . . . . . . . . . . . . . . . 420

6.1.2 Vector-Valued Analogues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426

6.1.3 Lp Estimates for Square Functions Associated

with Dyadic Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426

6.1.4 Lack of Orthogonality on Lp . . . . . . . . . . . . . . . . . . . . . . . . . . . 431

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434

6.2 Two Multiplier Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437

6.2.1 The Marcinkiewicz Multiplier Theorem on R . . . . . . . . . . . . . 439

6.2.2 The Marcinkiewicz Multiplier Theorem on Rn . . . . . . . . . . . . 441

6.2.3 The Mihlin–Hormander Multiplier Theorem on Rn . . . . . . . . 445

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450

6.3 Applications of Littlewood–Paley Theory . . . . . . . . . . . . . . . . . . . . . . 453

6.3.1 Estimates for Maximal Operators . . . . . . . . . . . . . . . . . . . . . . . 453

6.3.2 Estimates for Singular Integrals with Rough Kernels . . . . . . . 455

6.3.3 An Almost Orthogonality Principle on Lp . . . . . . . . . . . . . . . . 459

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461

6.4 The Haar System, Conditional Expectation, and Martingales . . . . . . 463

6.4.1 Conditional Expectation and Dyadic Martingale

Differences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464

6.4.2 Relation Between Dyadic Martingale Differences

and Haar Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465

6.4.3 The Dyadic Martingale Square Function . . . . . . . . . . . . . . . . . 469

6.4.4 Almost Orthogonality Between the Littlewood–Paley

Operators and the Dyadic Martingale Difference

Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474

6.5 The Spherical Maximal Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475

6.5.1 Introduction of the Spherical Maximal Function . . . . . . . . . . 475

6.5.2 The First Key Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478

6.5.3 The Second Key Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479

6.5.4 Completion of the Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481

6.6 Wavelets and Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482

6.6.1 Some Preliminary Facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483

6.6.2 Construction of a Nonsmooth Wavelet . . . . . . . . . . . . . . . . . . . 485

xvi Contents

6.6.3 Construction of a Smooth Wavelet . . . . . . . . . . . . . . . . . . . . . . 486

6.6.4 Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494

7 Weighted Inequalities 499

7.1 The Ap Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499

7.1.1 Motivation for the Ap Condition . . . . . . . . . . . . . . . . . . . . . . . . 500

7.1.2 Properties of Ap Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511

7.2 Reverse Holder Inequality for Ap Weights and Consequences . . . . . . 514

7.2.1 The Reverse Holder Property of Ap Weights . . . . . . . . . . . . . . 514

7.2.2 Consequences of the Reverse Holder Property . . . . . . . . . . . . 518

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521

7.3 The A∞ Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525

7.3.1 The Class of A∞ Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525

7.3.2 Characterizations of A∞ Weights . . . . . . . . . . . . . . . . . . . . . . . 527

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 530

7.4 Weighted Norm Inequalities for Singular Integrals . . . . . . . . . . . . . . . 532

7.4.1 Singular Integrals of Non Convolution type . . . . . . . . . . . . . . 532

7.4.2 A Good Lambda Estimate for Singular Integrals . . . . . . . . . . 533

7.4.3 Consequences of the Good Lambda Estimate . . . . . . . . . . . . . 539

7.4.4 Necessity of the Ap Condition . . . . . . . . . . . . . . . . . . . . . . . . . . 543

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545

7.5 Further Properties of Ap Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546

7.5.1 Factorization of Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546

7.5.2 Extrapolation from Weighted Estimates on a Single Lp0 . . . . 548

7.5.3 Weighted Inequalities Versus Vector-Valued Inequalities . . . . 554

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558

A Gamma and Beta Functions 563

A.1 A Useful Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563

A.2 Definitions of Γ (z) and B(z,w) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563

A.3 Volume of the Unit Ball and Surface of the Unit Sphere . . . . . . . . . . . 565

A.4 Computation of Integrals Using Gamma Functions . . . . . . . . . . . . . . . 565

A.5 Meromorphic Extensions of B(z,w) and Γ (z) . . . . . . . . . . . . . . . . . . . 566

A.6 Asymptotics of Γ (x) as x→ ∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567

A.7 Euler’s Limit Formula for the Gamma Function . . . . . . . . . . . . . . . . . 568

A.8 Reflection and Duplication Formulas for the Gamma Function . . . . . 570

B Bessel Functions 573

B.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573

B.2 Some Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573

B.3 An Interesting Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576

B.4 The Fourier Transform of Surface Measure on Sn−1 . . . . . . . . . . . . . . 577

B.5 The Fourier Transform of a Radial Function on Rn . . . . . . . . . . . . . . . 577

Contents xvii

B.6 Bessel Functions of Small Arguments . . . . . . . . . . . . . . . . . . . . . . . . . 578

B.7 Bessel Functions of Large Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . 579

B.8 Asymptotics of Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 580

B.9 Bessel Functions of general complex indices . . . . . . . . . . . . . . . . . . . . 582

C Rademacher Functions 585

C.1 Definition of the Rademacher Functions . . . . . . . . . . . . . . . . . . . . . . . . 585

C.2 Khintchine’s Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586

C.3 Derivation of Khintchine’s Inequalities . . . . . . . . . . . . . . . . . . . . . . . . 586

C.4 Khintchine’s Inequalities for Weak Type Spaces . . . . . . . . . . . . . . . . . 589

C.5 Extension to Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589

D Spherical Coordinates 591

D.1 Spherical Coordinate Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591

D.2 A Useful Change of Variables Formula . . . . . . . . . . . . . . . . . . . . . . . . 592

D.3 Computation of an Integral over the Sphere . . . . . . . . . . . . . . . . . . . . . 593

D.4 The Computation of Another Integral over the Sphere . . . . . . . . . . . . 593

D.5 Integration over a General Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594

D.6 The Stereographic Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594

E Some Trigonometric Identities and Inequalities 597

F Summation by Parts 599

G Basic Functional Analysis 601

H The Minimax Lemma 603

I Taylor’s and Mean Value Theorem in Several Variables 607

I.1 Mutlivariable Taylor’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 607

I.2 The Mean value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608

J The Whitney Decomposition of Open Sets in Rn 609

J.1 Decomposition of Open Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 609

J.2 Partition of Unity adapted to Whitney cubes . . . . . . . . . . . . . . . . . . . . 611

Glossary 613

References 617

Index 633

Chapter 1

Lp Spaces and Interpolation

Many quantitative properties of functions are expressed in terms of their integra-

bility to a power. For this reason it is desirable to acquire a good understanding

of spaces of functions whose modulus to a power p is integrable. These are called

Lebesgue spaces and are denoted by Lp. Although an in-depth study of Lebesgue

spaces falls outside the scope of this book, it seems appropriate to devote a chapter

to reviewing some of their fundamental properties.

The emphasis of this review is basic interpolation between Lebesgue spaces.

Many problems in Fourier analysis concern boundedness of operators on Lebesgue

spaces, and interpolation provides a framework that often simplifies this study. For

instance, in order to show that a linear operator maps Lp to itself for all 1 < p < ∞,

it is sufficient to show that it maps the (smaller) Lorentz space Lp,1 into the (larger)

Lorentz space Lp,∞ for the same range of p’s. Moreover, some further reductions can

be made in terms of the Lorentz space Lp,1. This and other considerations indicate

that interpolation is a powerful tool in the study of boundedness of operators.

Although we are mainly concerned with Lp subspaces of Euclidean spaces, we

discuss in this chapter Lp spaces of arbitrary measure spaces, since they represent

a useful general setting. Many results in the text require working with general mea-

sures instead of Lebesgue measure.

1.1 Lp and Weak Lp

A measure space is a set X equipped with a σ -algebra of subsets of it and a function

µ from the σ -algebra to [0,∞] that satisfies µ( /0) = 0 and

µ( ∞⋃

j=1

B j

)=

∞

∑j=1

µ(B j)

for any sequence B j of pairwise disjoint elements of the σ -algebra. The function µis called a (positive) measure on X and elements of the σ -algebra of X are called

L. Grafakos, Classical Fourier Analysis, Graduate Texts in Mathematics 249,

DOI 10.1007/978-1-4939-1194-3 1, © Springer Science+Business Media New York 2014

1

2 1 Lp Spaces and Interpolation

measurable sets. Measure spaces will be assumed to be complete, i.e., subsets of

the σ -algebra of measure zero also belong to the σ -algebra. A measure space X is

called σ -finite if there is a sequence of measurable subsets Xn of it such that

X =∞⋃

n=1

Xn

and µ(Xn)<∞. A real-valued function f on a measure space is called measurable if

the set x ∈ X : f (x)> λ is measurable for all real numbers λ . A complex-valued

function is measurable if and only if its real and imaginary parts are measurable. A

simple function is a finite linear combination of characteristic functions of measur-

able subsets of X ; these subsets may have infinite measure. A finitely simple function

has the formN

∑j=1

c jχB j

where N <∞, c j ∈C, and B j are pairwise disjoint measurable sets with µ(B j)<∞.

If N = ∞, this function will be called countably simple. Finitely simple functions

are exactly the integrable simple functions. Every nonnegative measurable function

is the pointwise limit of an increasing sequence of simple functions; if the space is

σ -finite, these simple functions can be chosen to be finitely simple.

For 0 < p < ∞, Lp(X ,µ) denotes the set of all complex-valued µ-measurable

functions on X whose modulus to the pth power is integrable. L∞(X ,µ) is the set of

all complex-valued µ-measurable functions f on X such that for some B > 0, the set

x : | f (x)|>B has µ-measure zero. Two functions in Lp(X ,µ) are considered equal

if they are equal µ-almost everywhere. When 0 < p < ∞ finitely simple functions

are dense in Lp(X ,µ). Within context and in the absence of ambiguity, Lp(X ,µ) is

simply written as Lp.

The notation Lp(Rn) is reserved for the space Lp(Rn, | · |), where | · | denotes n-

dimensional Lebesgue measure. Lebesgue measure on Rn is also denoted by dx.

Other measures will be considered on the Borel σ -algebra of Rn, i.e., is the smallest

σ -algebra that contains the closed subsets of Rn. Measures on the σ -algebra of Borel

measurable subsets are called Borel measures; such measures will be assumed to be

finite on compact subsets of Rn. A Borel measure µ with µ(Rn)<∞ is called a finite

Borel measure. A Borel measure on Rn is called regular for all Borel measurable

sets E we have

µ(E) = infµ(O) : E O, O open= supµ(K) : K E, K compact.

The space Lp(Z) equipped with counting measure is denoted by ℓp(Z) or simply ℓp.

For 0 < p <∞, we define the Lp norm of a function f (or quasi-norm if p < 1) by

∥∥ f∥∥

Lp(X ,µ)=

(∫

X| f (x)|p dµ(x)

) 1p

(1.1.1)

1.1 Lp and Weak Lp 3

and for p = ∞ by

∥∥ f∥∥

L∞(X ,µ)= ess.sup | f |= inf

B > 0 : µ(x : | f (x)|> B) = 0

. (1.1.2)

It is well known that Minkowski’s (or the triangle) inequality

∥∥ f +g∥∥

Lp(X ,µ)≤

∥∥ f∥∥

Lp(X ,µ)+

∥∥g∥∥

Lp(X ,µ)(1.1.3)

holds for all f , g in Lp = Lp(X ,µ), whenever 1 ≤ p ≤ ∞. Since in addition

‖ f‖Lp(X ,µ) = 0 implies that f = 0 (µ-a.e.), the Lp spaces are normed linear spaces

for 1≤ p≤∞. For 0 < p < 1, inequality (1.1.3) is reversed when f ,g≥ 0. However,

the following substitute of (1.1.3) holds:

∥∥ f +g∥∥

Lp(X ,µ)≤ 2

1−pp(∥∥ f

∥∥Lp(X ,µ)

+∥∥g

∥∥Lp(X ,µ)

), (1.1.4)

and thus Lp(X ,µ) is a quasi-normed linear space. See also Exercise 1.1.5. For all

0 < p ≤ ∞, it can be shown that every Cauchy sequence in Lp(X ,µ) is convergent,

and hence the spaces Lp(X ,µ) are complete. For the case 0 < p < 1 we refer to

Exercise 1.1.8. Therefore, the Lp spaces are Banach spaces for 1≤ p≤∞ and quasi-

Banach spaces for 0 < p < 1. For any p ∈ (0,∞)\1 we use the notation p′ = pp−1

.

Moreover, we set 1′ = ∞ and ∞′ = 1, so that p′′ = p for all p ∈ (0,∞]. Holder’s

inequality says that for all p ∈ [1,∞] and all measurable functions f ,g on (X ,µ) we

have ∥∥ f g∥∥

L1 ≤∥∥ f

∥∥Lp

∥∥g∥∥

Lp′ .

It is a well-known fact that the dual (Lp)∗ of Lp is isometric to Lp′ for all 1≤ p <∞.

Furthermore, the Lp norm of a function can be obtained via duality when 1≤ p≤∞as follows: ∥∥ f

∥∥Lp = sup

‖g‖Lp′=1

∣∣∣∣∫

Xf gdµ

∣∣∣∣ .

For the endpoint cases p = 1, p = ∞, see Exercise 1.4.12 (a), (b).

1.1.1 The Distribution Function

Definition 1.1.1. For f a measurable function on X , the distribution function of f is

the function d f defined on [0,∞) as follows:

d f (α) = µ(x ∈ X : | f (x)|> α) . (1.1.5)

The distribution function d f provides information about the size of f but not

about the behavior of f itself near any given point. For instance, a function on Rn and

each of its translates have the same distribution function. It follows from Definition

1.1.1 that d f is a decreasing function of α (not necessarily strictly).

f (x)

a3

a2

a1

E3 E1

B1

B2

B3

E2 x a1a2a30 0 α

αf ( )d

.

.

.

.

Fig. 1.1 The graph of a simple function f =∑3k=1 akχEk

and its distribution function d f (α). Here

B j =∑jk=1 µ(Ek).

Example 1.1.2. For pedagogical reasons we compute the distribution function d f of

a nonnegative simple function

f (x) =N

∑j=1

a jχE j(x) ,

where the sets E j are pairwise disjoint and a1 > · · ·> aN > 0. If α ≥ a1, then clearly

d f (α) = 0. However, if a2 ≤ α < a1 then | f (x)|> α precisely when x ∈ E1, and in

general, if a j+1 ≤ α < a j, then | f (x)|> α precisely when x ∈ E1∪·· ·∪E j. Setting

B j =j

∑k=1

µ(Ek) ,

for j ∈ 1, . . . ,N, B0 = aN+1 = 0, and a0 = ∞, we have

d f (α) =N

∑j=0

B jχ[a j+1,a j)(α) .

Note that these formulas are valid even when µ(Ei) = ∞ for some i. Figure 1.1

presents an illustration of this example when N = 3 and µ(E j)< ∞ for all j.

Proposition 1.1.3. Let f and g be measurable functions on (X ,µ). Then for all

α,β > 0 we have

(1) |g| ≤ | f | µ-a.e. implies that dg ≤ d f ;

(2) dc f (α) = d f (α/|c|), for all c ∈ C\0;(3) d f+g(α+β )≤ d f (α)+dg(β );

(4) d f g(αβ )≤ d f (α)+dg(β ).

Proof. The simple proofs are left to the reader.

Knowledge of the distribution function d f provides sufficient information to eval-

uate the Lp norm of a function f precisely. We state and prove the following impor-

tant description of the Lp norm in terms of the distribution function.

Proposition 1.1.4. Let (X ,µ) be a σ -finite measure space. Then for f in Lp(X ,µ),0 < p < ∞, we have ∥∥ f

∥∥p

Lp = p

∫ ∞

0α p−1d f (α)dα . (1.1.6)

Moreover, for any increasing continuously differentiable function ϕ on [0,∞) with

ϕ(0) = 0 and every measurable function f on X with ϕ(| f |) integrable on X, we

have ∫

Xϕ(| f |)dµ =

∫ ∞

0ϕ ′(α)d f (α)dα . (1.1.7)

Proof. Indeed, we have

p

∫ ∞

0α p−1d f (α)dα = p

∫ ∞

0α p−1

∫

Xχx: | f (x)|>α dµ(x)dα

=∫

X

∫ | f (x)|

0pα p−1 dα dµ(x)

=∫

X| f (x)|p dµ(x)

=∥∥ f

∥∥p

Lp ,

where in the second equality we used Fubini’s theorem, which requires the measure

space to be σ -finite. This proves (1.1.6). Identity (1.1.7) follows similarly, replacing

the function α p by the more general function ϕ(α) which has similar properties.

Definition 1.1.5. For 0 0 : d f (α)≤Cp

α pfor all α > 0

(1.1.8)

= supγ d f (γ)

1/p : γ > 0

(1.1.9)

is finite. The space weak L∞(X ,µ) is by definition L∞(X ,µ).

One should check that (1.1.9) and (1.1.8) are in fact equal. The weak Lp spaces are

denoted by Lp,∞(X ,µ). Two functions in Lp,∞(X ,µ) are considered equal if they are

equal µ-a.e. The notation Lp,∞(Rn) is reserved for Lp,∞(Rn, | · |). Using Proposition

1.1.3 (2), we can easily show that

∥∥k f∥∥

Lp,∞ = |k|∥∥ f

∥∥Lp,∞ , (1.1.10)

for any complex constant k. The analogue of (1.1.3) is

∥∥ f +g∥∥

Lp,∞ ≤ cp

(∥∥ f∥∥

Lp,∞ +∥∥g

∥∥Lp,∞

), (1.1.11)

where cp = max(2,21/p), a fact that follows from Proposition 1.1.3 (3), taking both

α and β equal to α/2. We also have that

∥∥ f∥∥

Lp,∞(X ,µ)= 0⇒ f = 0 µ-a.e. (1.1.12)

In view of (1.1.10), (1.1.11), and (1.1.12), Lp,∞ is a quasi-normed linear space for

0 < p < ∞.

The weak Lp spaces are larger than the usual Lp spaces. We have the following:

Proposition 1.1.6. For any 0 < p < ∞ and any f in Lp(X ,µ) we have

∥∥ f∥∥

Lp,∞ ≤∥∥ f

∥∥Lp .

Hence the embedding Lp(X ,µ) Lp,∞(X ,µ) holds.

Proof. This is just a trivial consequence of Chebyshev’s inequality:

α pd f (α)≤∫

x: | f (x)|>α| f (x)|p dµ(x)≤ ‖ f‖p

Lp .

Using (1.1.9) we obtain that ‖ f‖Lp,∞ ≤ ‖ f‖Lp .

The inclusion Lp Lp,∞ is strict. For example, on Rn with the usual Lebesgue

measure, let h(x) = |x|−np . Obviously, h is not in Lp(Rn) but h is in Lp,∞(Rn) with

‖h‖Lp,∞(Rn) = v1/pn , where vn is the measure of the unit ball of Rn.

It is not immediate from their definition that the weak Lp spaces are complete

with respect to the quasi-norm ‖ · ‖Lp,∞ . The completeness of these spaces is proved

in Theorem 1.4.11, but it is also a consequence of Theorem 1.1.13, proved in this

section.

1.1.2 Convergence in Measure

Next we discuss some convergence notions. The following notion is important in

probability theory.

Definition 1.1.7. Let f , fn, n = 1,2, . . . , be measurable functions on the measure

space (X ,µ). The sequence fn is said to converge in measure to f if for all ε > 0

there exists an n0 ∈ Z+ such that

n > n0 =⇒ µ(x ∈ X : | fn(x)− f (x)|> ε)< ε . (1.1.13)

Remark 1.1.8. The preceding definition is equivalent to the following statement:

For all ε > 0 limn→∞

µ(x ∈ X : | fn(x)− f (x)|> ε) = 0 . (1.1.14)

Clearly (1.1.14) implies (1.1.13). To see the converse given ε > 0, pick 0 < δ < εand apply (1.1.13) for this δ . There exists an n0 ∈ Z+ such that

µ(x ∈ X : | fn(x)− f (x)|> δ)< δ

holds for n > n0. Since

µ(x ∈ X : | fn(x)− f (x)|> ε)≤ µ(x ∈ X : | fn(x)− f (x)|> δ) ,

we conclude that

µ(x ∈ X : | fn(x)− f (x)|> ε)< δ

for all n > n0. Let n→ ∞ to deduce that

limsupn→∞

µ(x ∈ X : | fn(x)− f (x)|> ε)≤ δ . (1.1.15)

Since (1.1.15) holds for all 0 < δ < ε , (1.1.14) follows by letting δ → 0.

Convergence in measure is a weaker notion than convergence in either Lp or Lp,∞,

0 < p≤ ∞, as the following proposition indicates:

Proposition 1.1.9. Let 0 < p≤ ∞ and fn, f be in Lp,∞(X ,µ).

(1) If fn, f are in Lp and fn → f in Lp, then fn → f in Lp,∞.

(2) If fn → f in Lp,∞, then fn converges to f in measure.

Proof. Fix 0 0 we have

µ(x ∈ X : | fn(x)− f (x)|> ε)≤ 1

ε p

∫

X| fn− f |p dµ .

This shows that convergence in Lp implies convergence in weak Lp. The case p =∞is tautological.

Given ε > 0 find an n0 such that for n > n0, we have

∥∥ fn− f∥∥

Lp,∞ = supα>0

αµ(x ∈ X : | fn(x)− f (x)|> α)1p < ε

1p+1 .

Taking α = ε , we conclude that convergence in Lp,∞ implies convergence in mea-

sure.

Example 1.1.10. Note that there is no general converse of statement (2) in the pre-

ceding proposition. Fix 0 < p < ∞ and on [0,1] define the functions

fk, j = k1/pχ( j−1

k, j

k), k ≥ 1, 1≤ j ≤ k.

Consider the sequence f1,1, f2,1, f2,2, f3,1, f3,2, f3,3, . . .. Observe that

|x : fk, j(x)> 0|= 1/k .

Therefore, fk, j converges to 0 in measure. Likewise, observe that

∥∥ fk, j

∥∥Lp,∞ = sup

α>0

α|x : fk, j(x)> α|1/p ≥ supk≥1

(k−1/k)1/p

k1/p= 1 ,

which implies that fk, j does not converge to 0 in Lp,∞.

It turns out that every sequence convergent in Lp(X ,µ) or in Lp,∞(X ,µ) has a

subsequence that converges a.e. to the same limit.

Theorem 1.1.11. Let fn and f be complex-valued measurable functions on a mea-

sure space (X ,µ) and suppose that fn converges to f in measure. Then some subse-

quence of fn converges to f µ-a.e.

Proof. For all k = 1,2, . . . choose inductively nk such that

µ(x ∈ X : | fnk(x)− f (x)|> 2−k)< 2−k (1.1.16)

and such that n1 < n2 < · · ·< nk < · · · . Define the sets

Ak = x ∈ X : | fnk(x)− f (x)|> 2−k .

Equation (1.1.16) implies that

µ

( ∞⋃

k=m

Ak

)≤

∞

∑k=m

µ(Ak)≤∞

∑k=m

2−k = 21−m (1.1.17)

for all m = 1,2,3, . . . . It follows from (1.1.17) that

µ

( ∞⋃

k=1

Ak

)≤ 1 < ∞ . (1.1.18)

Using (1.1.17) and (1.1.18), we conclude that the sequence of the measures of the sets

⋃∞k=m Ak∞m=1 converges as m→ ∞ to

µ

( ∞⋂

m=1

∞⋃

k=m

Ak

)= 0 . (1.1.19)

To finish the proof, observe that the null set in (1.1.19) contains the set of all x ∈ X

for which fnk(x) does not converge to f (x).

In many situations we are given a sequence of functions and we would like to

extract a convergent subsequence. One way to achieve this is via the next theorem,

which is a useful variant of Theorem 1.1.11. We first give a relevant definition.

Definition 1.1.12. We say that a sequence of measurable functions fn on the mea-

sure space (X ,µ) is Cauchy in measure if for every ε > 0, there exists an n0 ∈ Z+

such that for n,m > n0 we have

µ(x ∈ X : | fm(x)− fn(x)|> ε)< ε .

Theorem 1.1.13. Let (X ,µ) be a measure space and let fn be a complex-valued

sequence on X that is Cauchy in measure. Then some subsequence of fn converges

µ-a.e.

Proof. The proof is very similar to that of Theorem 1.1.11. For all k = 1,2, . . . choose

nk inductively such that

µ(x ∈ X : | fnk(x)− fnk+1

(x)|> 2−k)< 2−k (1.1.20)

and such that n1 < n2 < · · ·< nk < nk+1 < · · · . Define

Ak = x ∈ X : | fnk(x)− fnk+1

(x)|> 2−k .

As shown in the proof of Theorem 1.1.11, (1.1.20) implies that

µ

( ∞⋂

m=1

∞⋃

k=m

Ak

)= 0 . (1.1.21)

For x /∈⋃∞k=m Ak and i≥ j ≥ j0 ≥ m (and j0 large enough) we have

| fni(x)− fn j

(x)| ≤i−1

∑l= j

| fnl(x)− fnl+1

(x)| ≤i−1

∑l= j

2−l ≤ 21− j ≤ 21− j0 .

This implies that the sequence fni(x)i is Cauchy for every x in the set (

⋃∞k=m Ak)

c

and therefore converges for all such x. We define a function

f (x) =

⎧⎨⎩

limj→∞

fn j(x) when x /∈⋂∞

m=1

⋃∞k=m Ak ,

0 when x ∈⋂∞m=1

⋃∞k=m Ak .

Then fn j→ f almost everywhere.

1.1.3 A First Glimpse at Interpolation

It is a useful fact that if a function f is in Lp(X ,µ) and in Lq(X ,µ), then it also lies

in Lr(X ,µ) for all p < r < q. The usefulness of the spaces Lp,∞ can be seen from the

following sharpening of this statement:

Proposition 1.1.14. Let 0 < p < q≤ ∞ and let f in Lp,∞(X ,µ)∩Lq,∞(X ,µ), where

X is a σ -finite measure space. Then f is in Lr(X ,µ) for all p < r < q and

∥∥ f∥∥

Lr ≤(

r

r− p+

r

q− r

)1r ∥∥ f

∥∥1r − 1

q1p− 1

q

Lp,∞

∥∥ f∥∥

1p− 1

r1p− 1

q

Lq,∞ , (1.1.22)

with the interpretation that 1/∞= 0.

Proof. Let us take first q < ∞. We know that

d f (α)≤min

(∥∥ f∥∥p

Lp,∞

α p,

∥∥ f∥∥q

Lq,∞

αq

). (1.1.23)

Set

B =

(∥∥ f∥∥q

Lq,∞∥∥ f∥∥p

Lp,∞

) 1q−p

. (1.1.24)

We now estimate the Lr norm of f . By (1.1.23), (1.1.24), and Proposition 1.1.4 we

have

∥∥ f∥∥r

Lr(X ,µ)= r

∫ ∞

0αr−1d f (α)dα

≤ r

∫ ∞

0αr−1 min

(∥∥ f∥∥p

Lp,∞

α p,

∥∥ f∥∥q

Lq,∞

αq

)dα

= r

∫ B

0αr−1−p

∥∥ f∥∥p

Lp,∞ dα+ r

∫ ∞

Bαr−1−q

∥∥ f∥∥q

Lq,∞ dα

=r

r− p

∥∥ f∥∥p

Lp,∞Br−p +r

q− r

∥∥ f∥∥q

Lq,∞Br−q

=

(r

r− p+

r

q− r

)(∥∥ f∥∥p

Lp,∞

) q−rq−p

(∥∥ f∥∥q

Lq,∞

) r−pq−p .

(1.1.25)

Observe that the integrals converge, since r− p > 0 and r−q < 0.

The case q = ∞ is easier. Since d f (α) = 0 for α > ‖ f‖L∞ we need to use only

the inequality d f (α)≤ α−p‖ f‖pLp,∞ for α ≤ ‖ f‖L∞ in estimating the first integral in

(1.1.25). We obtain ∥∥ f∥∥r

Lr ≤r

r− p

∥∥ f∥∥p

Lp,∞

∥∥ f∥∥r−p

L∞,

which is nothing other than (1.1.22) when q = ∞. This completes the proof.

Note that (1.1.22) holds with constant 1 if Lp,∞ and Lq,∞ are replaced by Lp and

Lq, respectively. It is often convenient to work with functions that are only locally in

some Lp space. This leads to the following definition.

Definition 1.1.15. For 0 < p < ∞, the space Lploc(R

n, | · |) or simply Lploc(R

n) is the

set of all Lebesgue-measurable functions f on Rn that satisfy

∫

K| f (x)|p dx < ∞ (1.1.26)

for any compact subset K of Rn. Functions that satisfy (1.1.26) with p = 1 are called

locally integrable functions on Rn.

The union of all Lp(Rn) spaces for 1 ≤ p ≤ ∞ is contained in L1loc(R

n). More

generally, for 0 0 and p < n/(n+α), and

observe that f is not integrable over any open set in Rn containing the origin.

Exercises

1.1.1. Suppose f and fn are measurable functions on (X ,µ). Prove that

(a) d f is right continuous on [0,∞).(b) If | f | ≤ liminfn→∞ | fn| µ-a.e., then d f ≤ liminfn→∞ d fn .

(c) If | fn| ↑ | f |, then d fn ↑ d f .[Hint: Part (a): Let tn be a decreasing sequence of positive numbers that tends to

zero. Show that d f (α0 + tn) ↑ d f (α0) using a convergence theorem. Part (b): Let

E = x ∈ X : | f (x)|> α and En = x ∈ X : | fn(x)|> α. Use that µ(⋂∞

n=m En

)≤

liminfn→∞

µ(En) and E ⋃∞

m=1

⋂∞n=m En µ-a.e.

]

1.1.2. (Holder’s inequality) Let 0 < p, p1, . . . , pk ≤ ∞, where k ≥ 2, and let f j be in

Lp j = Lp j(X ,µ). Assume that

1

p=

1

p1+ · · ·+ 1

pk

.

(a) Show that the product f1 · · · fk is in Lp and that

∥∥ f1 · · · fk

∥∥Lp ≤

∥∥ f1

∥∥Lp1· · ·

∥∥ fk

∥∥Lpk

.

(b) When no p j is infinite, show that if equality holds in part (a), then it must be the

case that c1| f1|p1 = · · ·= ck| fk|pk µ-a.e. for some c j ≥ 0.

(c) Let 0 < q < 1 and q′ = qq−1

. For r < 0 and g > 0 almost everywhere, define

‖g‖Lr = ‖g−1‖−1

L|r|. Show that if g is strictly positive µ-a.e. and lies in Lq′ and f is

measurable such that f g belongs to L1, we have

∥∥ f g∥∥

L1 ≥∥∥ f

∥∥Lq

∥∥g∥∥

Lq′ .

1.1.3. Let (X ,µ) be a measure space.

(a) If f is in Lp0(X ,µ) for some p0 < ∞, prove that

limp→∞

∥∥ f∥∥

Lp =∥∥ f

∥∥L∞

.

(b) (Jensen’s inequality) Suppose that µ(X) = 1. Show that

∥∥ f∥∥

Lp ≥ exp

(∫

Xlog | f (x)|dµ(x)

)

for all 0 0, then

limp→0

∥∥ f∥∥

Lp = exp

(∫

Xlog | f (x)|dµ(x)

)

with the interpretation e−∞ = 0.[Hint: Part (a): If 0 < ‖ f‖L∞ < ∞, use that ‖ f‖Lp ≤ ‖ f‖(p−p0)/p

L∞ ‖ f‖p0/p

Lp0to obtain

limsupp→∞ ‖ f‖Lp ≤ ‖ f‖L∞ . Conversely, let Eγ = x ∈ X : | f (x)|> γ‖ f‖L∞ for γ in

(0,1). Then µ(Eγ) > 0, ‖ f‖Lp0 (Eγ ) > 0, and ‖ f‖Lp ≥(γ‖ f‖L∞

)(p−p0)/p‖ f‖p0/p

Lp0 (Eγ ),

hence liminfp→∞ ‖ f‖Lp ≥ γ‖ f‖L∞ . If ‖ f‖L∞ = ∞, set Gn = | f | > n and use that

‖ f‖Lp ≥ ‖ f‖Lp(Gn) ≥ nµ(Gn)1p to obtain liminfp→∞ ‖ f‖Lp ≥ n. Part (b) is a direct

consequence of Jensen’s inequality∫

X log |h|dµ ≤ log(∫

X |h|dµ). Part (c): Fix a

sequence 0 < pn < p0 such that pn ↓ 0 and define

hn(x) =1

p0(| f (x)|p0 −1)− 1

pn

(| f (x)|pn −1).

Use that 1p(t p−1) ↓ log t as p ↓ 0 for all t > 0. The Lebesgue monotone convergence

theorem yields∫

X hn dµ ↑ ∫X hdµ , hence

∫X

1pn(| f |pn −1)dµ ↓ ∫

X log | f |dµ , where

the latter could be −∞. Use

exp

(∫

Xlog | f |dµ

)≤

(∫

X| f |pn dµ

) 1pn

≤ exp

(∫

X

1

pn

(| f |pn −1)dµ

)

to complete the proof.]

1.1.4. Let a j be a sequence of positive reals. Show that

(a)(∑∞j=1 a j

)θ ≤ ∑∞j=1 aθj , for any 0≤ θ ≤ 1.

(b) ∑∞j=1 aθj ≤(∑∞j=1 a j

)θ, for any 1≤ θ < ∞.

(c)(∑N

j=1 a j

)θ ≤ Nθ−1∑Nj=1 aθj , when 1≤ θ < ∞.

(d) ∑Nj=1 aθj ≤ N1−θ(∑N

j=1 a j

)θ, when 0≤ θ ≤ 1.

1.1.5. Let f jNj=1 be a sequence of Lp(X ,µ) functions.

(a) (Minkowski’s inequality) For 1≤ p≤ ∞ show that

∥∥N

∑j=1

f j

∥∥Lp ≤

N

∑j=1

∥∥ f j

∥∥Lp .

(b) (Reverse Minkowski inequality) For 0 < p < 1 and f j ≥ 0 prove that

N

∑j=1

∥∥ f j

∥∥Lp ≤

∥∥N

∑j=1

f j

∥∥Lp .

(c) For 0 < p < 1 show that

∥∥N

∑j=1

f j

∥∥Lp ≤ N

1−pp

N

∑j=1

∥∥ f j

∥∥Lp .

(d) The constant N1−p

p in part (c) is best possible.[Hint: Part (c): Use Exercise 1.1.4 (c). Part (d): Take f jN

j=1 to be characteristic

functions of disjoint sets with the same measure.]

1.1.6. (a) (Minkowski’s integral inequality) Let (X ,µ) and (T,ν) be two σ -finite

measure spaces and let 1 ≤ p < ∞. Show that for every nonnegative measurable

function F on the product space (X ,µ)× (T,ν) we have

[∫

T

(∫

XF(x, t)dµ(x)

)p

dν(t)

] 1p

≤∫

X

[∫

TF(x, t)p dν(t)

] 1p

dµ(x) ,

(b) State and prove an analogous inequality when p = ∞.

(c) Prove that when 0 < p < 1, then the preceding inequality is reversed.

(d) (Y. Sawano) Consider the example X = T = [0,1], µ is counting measure, ν is

Lebesgue measure, F(x, t) = 1 when x = t and zero otherwise. What is the relevance

of this example with the inequalities in (a) and (b)?[Hint: Part (a) Split the power p as 1+(p− 1) and apply Holder’s inequality with

exponents p and p′. Part (b) Let p→ ∞ on subsets of X with finite measure.]

1.1.7. Let f1, . . . , fN be in Lp,∞(X ,µ).(a) Prove that for 1≤ p < ∞ we have

∥∥N

∑j=1

f j

∥∥Lp,∞ ≤ N

N

∑j=1

∥∥ f j

∥∥Lp,∞ .

(b) Show that for 0 α) ≤ ∑N

j=1 µ(| f j| > α/N) and Exercise

1.1.4 (a) and (c).]

1.1.8. Let 0 < p < ∞. Prove that Lp(X ,µ) is a complete quasi-normed space. This

means that every quasi-norm Cauchy sequence is quasi-norm convergent.

[Hint: Let fn be a Cauchy sequence in Lp. Pass to a subsequence nii such that

‖ fni+1− fni

‖Lp ≤ 2−i. Then the series f = fn1+∑∞i=1( fni+1

− fni) converges in Lp.

]

1.1.9. Let (X ,µ) be a measure space with µ(X) < ∞. Suppose that a sequence of

measurable functions fn on X converges to f µ-a.e. Prove that fn converges to f in

measure.[Hint: For ε > 0,

x ∈ X : fn(x)→ f (x)

∞⋃m=1

∞⋂n=mx ∈ X : | fn(x)− f (x)|< ε

.]

1.1.10. Let f be a measurable function on (X ,µ) such such d f (α)<∞ for all α > 0.

Fix γ > 0 and define fγ = f χ| f |>γ and f γ = f − fγ = f χ| f |≤γ .(a) Prove that

d fγ (α) =

d f (α) when α > γ ,

d f (γ) when α ≤ γ ,

d f γ (α) =

0 when α ≥ γ ,

d f (α)−d f (γ) when α < γ .

(b) If f ∈ Lp(X ,µ) then

∥∥ fγ∥∥p

Lp = p

∫ ∞

γα p−1d f (α)dα+ γ pd f (γ),

∥∥ f γ∥∥p

Lp = p

∫ γ

0α p−1d f (α)dα− γ pd f (γ),

∫

γ<| f |≤δ| f |p dµ = p

∫ δ

γd f (α)α

p−1 dα−δ pd f (δ )+ γ pd f (γ).

(c) If f is in Lp,∞(X ,µ) prove that f γ is in Lq(X ,µ) for any q > p and fγ is in

Lq(X ,µ) for any q < p. Thus Lp,∞ Lp0 +Lp1 when 0 < p0 < p < p1 ≤ ∞.

1.1.11. Let (X ,µ) be a measure space and let E be a subset of X with µ(E) < ∞.

Assume that f is in Lp,∞(X ,µ) for some 0 < p < ∞.

(a) Show that for 0 < q α

)≤min

(µ(E),α−p

∥∥ f∥∥p

Lp,∞

).]

1.1.12. (Normability of weak Lp for p > 1) Let (X ,µ) be a σ -finite measure space

and let 0 < p < ∞. Pick 0 < r < p and define

f

Lp,∞ = sup0<µ(E)<∞

µ(E)−1r +

1p

(∫

E| f |rdµ

) 1r

,

where the supremum is taken over all measurable subsets E of X of finite measure.

(a) Use Exercise 1.1.11 with q = r to conclude that

f

Lp,∞ ≤(

p

p− r

) 1r ∥∥ f

∥∥Lp,∞

for all f in Lp,∞(X ,µ). (It is not needed that X be σ -finite here).

(b) Prove that for all f in Lp,∞(X ,µ) we have

∥∥ f∥∥

Lp,∞ ≤ f

Lp,∞ .

(Y. Oi) Notice that if X = 1,2, µ(1) = 1, µ(2) = ∞, then X is not σ -finite,

and verify that for the function f = 1 the preceding inequality fails.

(c) Show that Lp,∞(X ,µ) is metrizable for all 0 1, i.e., there is a norm on the space equivalent to ‖ · ‖Lp,∞ .

(d) Use the characterization of the weak Lp quasi-norm obtained in parts (a) and (b)

to prove Fatou’s lemma for this space: For all measurable functions gn on X we have

∥∥ liminfn→∞

|gn|∥∥

Lp,∞ ≤Cp liminfn→∞

∥∥gn

∥∥Lp,∞

for some constant Cp that depends only on p ∈ (0,∞).[Hint: Part (b): Write X =

⋃∞k=1 Xk with µ(Xk)< ∞ and take E = | f |> α∩Xk.

]

1.1.13. Consider the N! functions on the line

fσ =N

∑j=1

N

σ( j)χ[ j−1

N , jN )

,

where σ is a permutation of the set 1,2, . . . ,N.(a) Show that each fσ satisfies ‖ fσ‖L1,∞ = 1.

(b) Show that ‖∑σ∈SNfσ‖L1,∞ = N!

(1+ 1

2+ · · ·+ 1

N

).

(c) Conclude that the space L1,∞(R) is not normable (this means that ‖ · ‖L1,∞ is not

equivalent to a norm).

(d) Use a similar argument to prove that L1,∞(Rn) is not normable by considering

the functions

fσ (x1, . . . ,xn) =N

∑j1=1

· · ·N

∑jn=1

Nn

σ(τ( j1, . . . , jn))χ[

j1−1

N ,j1N )(x1) · · ·χ[ jn−1

N , jnN )(xn) ,

where σ is a permutation of the set 1,2, . . . ,Nn and τ is a fixed injective map

from the set of all n-tuples of integers with coordinates 1 ≤ j ≤ N onto the set

1,2, . . . ,Nn. One may take

τ( j1, . . . , jn) = j1 +N( j2−1)+N2( j3−1)+ · · ·+Nn−1( jn−1),

for instance.

1.1.14. Let (X ,µ) be a measure space and let s > 0.

(a) Let f be a measurable function on X . Show that if 0 < p < q < ∞ we have

∫

| f |≤s| f |q dµ ≤ q

q− psq−p

∥∥ f∥∥p

Lp,∞ .

(b) Let f j, 1≤ j ≤ m, be measurable functions on X and let 0 < p < ∞. Show that

∥∥∥ max1≤ j≤m

| f j|∥∥∥

p

Lp,∞≤

m

∑j=1

∥∥ f j

∥∥p

Lp,∞ .

(c) Conclude from part (b) that for 0 < p < 1 we have

∥∥ f1 + · · ·+ fm

∥∥p

Lp,∞ ≤2− p

1− p

m

∑j=1

∥∥ f j

∥∥p

Lp,∞ .

The latter estimate is referred to as the p-normability of weak Lp.[Hint: Part (a): Use the distribution function. Part (c): First obtain the estimate

d f1+···+ fm(α) ≤ µ(| f1+· · ·+ fm|>α,max | f j|≤α)+dmax j | f j |(α)

for all α > 0 and then use part (b).]

1.1.15. (Holder’s inequality for weak spaces) Let f j be in Lp j ,∞ of a measure space

X where 0 < p j < ∞ and 1≤ j ≤ k. Let

1

p=

1

p1+ · · ·+ 1

pk

.

Prove that∥∥ f1 · · · fk

∥∥Lp,∞ ≤ p

− 1p

k

∏j=1

p

1p j

j

k

∏j=1

∥∥ f j

∥∥L

p j ,∞ .

[Hint: Take ‖ f j‖L

p j ,∞ = 1 for all j. Control d f1··· fk(α) by

µ(| f1|>α/s1)+ · · ·+µ(| fk−1|>sk−2/sk−1)+µ(| fk|>sk−1)≤ (s1/α)

p1 +(s2/s1)p2 + · · ·+(sk−1/sk−2)

pk−1 +(1/sk−1)pk .

Set x1 = s1/α , x2 = s2/s1, . . . ,xk = 1/sk−1. Minimize xp11 + · · ·+ x

pk

k subject to the

constraint x1 · · ·xk = 1/α .]

1.1.16. Let 0 < p0 < p < p1 ≤ ∞ and let 1p= 1−θ

p0+ θ

p1for some θ ∈ [0,1]. Prove

the following:

∥∥ f∥∥

Lp ≤∥∥ f

∥∥1−θLp0

∥∥ f∥∥θ

Lp1,

∥∥ f∥∥

Lp,∞ ≤∥∥ f

∥∥1−θLp0 ,∞

∥∥ f∥∥θ

Lp1 ,∞.

1.2 Convolution and Approximate Identities 17

1.1.17. ([231]) Follow the steps below to prove the isoperimetric inequality. For

n ≥ 2 and 1 ≤ j ≤ n define the projection maps π j : Rn → Rn−1 by setting for

x = (x1, . . . ,xn),π j(x) = (x1, . . . ,x j−1,x j+1, . . . ,xn) ,

with the obvious interpretations when j = 1 or j = n.

(a) For maps f j : Rn−1 → C prove that

Λ( f1, . . . , fn) =∫

Rn

n

∏j=1

∣∣ f j π j

∣∣dx≤n

∏j=1

∥∥ f j

∥∥Ln−1(Rn−1)

.

(b) Let Ω be a compact set with a rectifiable boundary in Rn where n ≥ 2. Show

that there is a constant cn independent of Ω such that

|Ω | ≤ cn|∂Ω |n

n−1 ,

where the expression |∂Ω | denotes the (n−1)-dimensional surface measure of the

boundary of Ω .[Hint: Part (a): Use induction starting with n = 2. For n≥ 3 write

Λ( f1, . . . , fn) ≤∫

Rn−1P(x1, . . . ,xn−1)| fn(πn(x))|dx1 · · ·dxn−1

≤ ‖P‖L

n−1n−2 (Rn−1)

∥∥ fn πn

∥∥Ln−1(Rn−1)

,

where P(x1, . . . ,xn−1) =∫

R | f1(π1(x)) · · · fn−1(πn−1(x))|dxn, and apply the induc-

tion hypothesis to the n−1 functions

[∫

Rf j(π j(x))

n−1 dxn

] 1n−2

,

for j = 1, . . . ,n−1, to obtain the required conclusion. Part (b): Specialize part (a) to

the case f j = χπ j [Ω ] to obtain

|Ω | ≤ |π1[Ω ]| 1n−1 · · · |πn[Ω ]| 1

n−1

and then use that |π j[Ω ]| ≤ 12|∂Ω |.

]

1.2 Convolution and Approximate Identities

The notion of convolution can be defined on measure spaces endowed with a group

structure. It turns out that the most natural environment to define convolution is the

context of topological groups. Although the focus of this book is harmonic analysis

on Euclidean spaces, we develop the notion of convolution on general groups. This

allows us to study this concept on Rn, Zn, and Tn, in a unified way. Moreover,


since the basic properties of convolutions and approximate identities do not require

commutativity of the group operation, we may assume that the underlying groups

are not necessarily abelian. Thus, the results in this section can be also applied to

nonabelian structures such as the Heisenberg group.

1.2.1 Examples of Topological Groups

A topological group G is a Hausdorff topological space that is also a group with law

(x,y) → xy (1.2.1)

such that the maps (x,y) → xy and x → x−1 are continuous. The identity element of

the group is the unique element e with the property xe = ex = x for all x ∈ G. We

adopt the standard notation

AB = ab : a ∈ A,b ∈ B, A−1 = a−1 : a ∈ A

for subsets A and B of G. Note that (AB)−1 = B−1A−1. Every topological group

G has an open basis at e consisting of symmetric neighborhoods, i.e., open sets U

satisfying U = U−1. A topological group is called locally compact if there is an

open set U containing the identity element such that U is compact. Then every point

in the group has an open neighborhood with compact closure.

Let G be a locally compact group. It is known that G possesses a positive measure

λ on the Borel sets that is nonzero on all nonempty open sets, finite on compact sets,

and is left invariant, meaning that

λ (tA) = λ (A), (1.2.2)

for all measurable sets A and all t ∈ G. Such a measure λ is called a (left) Haar

measure on G. Similarly, G possesses a right Haar measure which is right invariant,

i.e., λ (At) = λ (A) for all measurable A G and all t ∈G. For the existence of Haar

measure we refer to [152, §15] or [213, §16.3]. Furthermore, Haar measure is unique

up to positive multiplicative constants. If G is abelian then any left Haar measure on

G is a constant multiple of any given right Haar measure on G. A locally compact

group which is a countable union of compact subsets is a σ -finite measure space

under left or right Haar measure. This is case for connected locally compact groups.

Example 1.2.1. The standard examples are provided by the spaces Rn and Zn with

the usual topology and the usual addition of n-tuples. Another example is the space

Tn = Rn/Zn defined as follows:

Tn = [0,1)×·· ·× [0,1)︸︷︷︸n times

with the usual topology and group law:

(x1, . . . ,xn)+(y1, . . . ,yn) = ((x1 + y1) mod1, . . . ,(xn + yn) mod1).

Example 1.2.2. Let G = R∗ = R\0 with group law the usual multiplication. It is

easy to verify that the measure λ = dx/|x| is invariant under multiplicative transla-

tions, that is, ∫ ∞

−∞f (tx)

dx

|x| =∫ ∞

−∞f (x)

dx

|x| ,

for all f in L1(G,µ) and all t ∈ R∗. Therefore, dx/|x| is a Haar measure. [Taking

f = χA gives λ (tA) = λ (A).]

Example 1.2.3. Similarly, on the multiplicative group G = R+, a Haar measure is

dx/x.

Example 1.2.4. Counting measure is a Haar measure on the group Zn with the usual

addition as group operation.

Example 1.2.5. The Heisenberg group Hn is the set Cn×R with the group operation

(z1, . . . ,zn, t)(w1, . . . ,wn,s) =(

z1 +w1, . . . ,zn +wn, t + s+2Imn

∑j=1

z jw j

).

It can easily be seen that the identity element e of this group is 0 ∈ Cn×R and

(z1, . . . ,zn, t)−1 = (−z1, . . . ,−zn,−t). Topologically the Heisenberg group is identi-

fied with Cn×R, and both left and right Haar measure on Hn is Lebesgue measure.

The norm

|(z1, . . . ,zn, t)|=[( n

∑j=1

|z j|2)2

+ t2

] 14

introduces balls Br(x)= y∈Hn : |y−1x|< r on the Heisenberg group that are quite

different from Euclidean balls. For x close to the origin, the balls Br(x) are not far

from being Euclidean, but for x far away from e = 0 they look like slanted truncated

cylinders. The Heisenberg group can be naturally identified as the boundary of the

unit ball in Cn and plays an important role in quantum mechanics.

1.2.2 Convolution

Throughout the rest of this section, we fix a locally compact group G and a left

invariant Haar measure λ on G. We assume that G is a countable union of compact

subsets, hence the pair (G,λ ) forms a σ -finite measure space. The spaces Lp(G,λ )and Lp,∞(G,λ ) are simply denoted by Lp(G) and Lp,∞(G).

Left invariance of λ is equivalent to the fact that for all t ∈G and all nonnegative

measurable functions f on G we have

∫

Gf (tx)dλ (x) =

∫

Gf (x)dλ (x) . (1.2.3)

Equation (1.2.3) is a restatement of (1.2.2) if f is a characteristic function. Obviously

(1.2.3) also holds for f ∈ L1(G) by linearity and approximation.

We are now ready to define the operation of convolution.

Definition 1.2.6. Let f , g be in L1(G). Define the convolution f ∗g by

( f ∗g)(x) =∫

Gf (y)g(y−1x)dλ (y) . (1.2.4)

For instance, if G = Rn with the usual additive structure, then y−1 = −y and the

integral in (1.2.4) is written as

( f ∗g)(x) =∫

Rnf (y)g(x− y)dy .

Remark 1.2.7. The right-hand side of (1.2.4) is defined a.e., since the following

double integral converges absolutely:

∫

G

∫

G| f (y)||g(y−1x)|dλ (y)dλ (x)

=∫

G

∫

G| f (y)||g(y−1x)|dλ (x)dλ (y)

=∫

G| f (y)|

∫

G|g(y−1x)|dλ (x)dλ (y)

=

∫

G| f (y)|

∫

G|g(x)|dλ (x)dλ (y) by (1.2.2)

=∥∥ f

∥∥L1(G)

∥∥g∥∥

L1(G)<+∞ .

The change of variables z = x−1y yields that (1.2.4) is in fact equal to

( f ∗g)(x) =∫

Gf (xz)g(z−1)dλ (z) , (1.2.5)

where the substitution of dλ (y) by dλ (z) is justified by left invariance.

Example 1.2.8. On R let f (x) = 1 when −1 ≤ x ≤ 1 and zero otherwise. We see

that ( f ∗ f )(x) is equal to the length of the intersection of the intervals [−1,1] and

[x− 1,x + 1]. It follows that ( f ∗ f )(x) = 2− |x| for |x| ≤ 2 and zero otherwise.

Observe that f ∗ f is a smoother function than f . Similarly, we obtain that f ∗ f ∗ f

is a smoother function than f ∗ f .

There is an analogous calculation when g is the characteristic function of the unit

disk B(0,1) in R2. A simple computation gives

(g∗g)(x) =∣∣B

(0,1

)∩B

(x,1

)∣∣=∫ +

√1− 1

4 |x|2

−√

1− 14 |x|2

(2√

1− t2−|x|)

dt

= 2arcsin

(√1− 1

4|x|2

)−|x|

√1− 1

4|x|2

when x = (x1,x2) in R2 satisfies |x| ≤ 2, while (g∗g)(x) = 0 if |x| ≥ 2.

A calculation similar to that in Remark 1.2.7 yields that

∥∥ f ∗g∥∥

L1(G)≤

∥∥ f∥∥

L1(G)

∥∥g∥∥

L1(G), (1.2.6)

that is, the convolution of two integrable functions is also an integrable function

with L1 norm less than or equal to the product of the L1 norms.

Proposition 1.2.9. For all f , g, h in L1(G), the following properties are valid:

(1) f ∗ (g∗h) = ( f ∗g)∗h (associativity)

(2) f ∗ (g+h) = f ∗g+ f ∗h and ( f +g)∗h = f ∗h+g∗h (distributivity)

Proof. The easy proofs are omitted.

Proposition 1.2.9 implies that L1(G) is a (not necessarily commutative) Banach

algebra under the convolution product.

1.2.3 Basic Convolution Inequalities

The most fundamental inequality involving convolutions is the following.

Theorem 1.2.10. (Minkowski’s inequality) Let 1≤ p≤ ∞. For f in Lp(G) and g in

L1(G) we have that g∗ f exists λ -a.e. and satisfies

∥∥g∗ f∥∥

Lp(G)≤

∥∥g∥∥

L1(G)

∥∥ f∥∥

Lp(G). (1.2.7)

Proof. Estimate (1.2.7) follows directly from Exercise 1.1.6. Here we give a direct

proof. We may assume that 1 < p < ∞, since the cases p = 1 and p = ∞ are simple.

We first show that the convolution |g| ∗ | f | exists λ -a.e. Indeed,

(|g| ∗ | f |)(x) =∫

G| f (y−1x)| |g(y)|dλ (y) . (1.2.8)

Apply Holder’s inequality in (1.2.8) with respect to the measure |g(y)|dλ (y) to the

functions y → f (y−1x) and 1 with exponents p and p′ = p/(p−1), respectively. We

obtain

(|g| ∗ | f |)(x)≤(∫

G| f (y−1x)|p|g(y)|dλ (y)

)1p(∫

G|g(y)|dλ (y)

)1p′. (1.2.9)

Taking Lp norms of both sides of (1.2.9) we deduce

∥∥|g| ∗ | f |∥∥

Lp ≤(∥∥g

∥∥p−1

L1

∫

G

∫

G| f (y−1x)|p|g(y)|dλ (y)dλ (x)

)1p

=

(∥∥g∥∥p−1

L1

∫

G

∫

G| f (y−1x)|p dλ (x)|g(y)|dλ (y)

)1p

=

(∥∥g∥∥p−1

L1

∫

G

∫

G| f (x)|p dλ (x)|g(y)|dλ (y)

)1p

by (1.2.3)

=

(∥∥ f∥∥p

Lp

∥∥g∥∥

L1

∥∥g∥∥p−1

L1

)1p

=∥∥ f

∥∥Lp

∥∥g∥∥

L1 < ∞ ,

where the second equality follows by Fubini’s theorem. This shows that |g| ∗ | f | is

finite λ -a.e. and satisfies (1.2.7); then g ∗ f exists λ -a.e. and also satisfies (1.2.7),

since |g∗ f | ≤ |g| ∗ | f |.

Remark 1.2.11. Theorem 1.2.10 may fail for nonabelian groups if g ∗ f is replaced

by f ∗g in (1.2.7). Note, however, that if for all h ∈ L1(G) we have

∥∥h∥∥

L1 =∥∥h

∥∥L1 , (1.2.10)

where h(x) = h(x−1), then (1.2.7) holds when the quantity ‖g ∗ f‖Lp(G) is replaced

by ‖ f ∗g‖Lp(G). To see this, observe that if (1.2.10) holds, then we can use (1.2.5) to

conclude that if f in Lp(G) and g in L1(G), then

∥∥ f ∗g∥∥

Lp(G)≤

∥∥g∥∥

L1(G)

∥∥ f∥∥

Lp(G). (1.2.11)

If the left Haar measure satisfies

λ (A) = λ (A−1) (1.2.12)

for all measurable A G, then (1.2.10) holds and thus (1.2.11) is satisfied for all g in

L1(G) and f ∈ Lp(G). This is, for instance, the case for the Heisenberg group Hn.

Minkowski’s inequality (1.2.11) is only a special case of Young’s inequality in

which the function g can be in any space Lr(G) for 1≤ r ≤ ∞.

Theorem 1.2.12. (Young’s inequality) Let 1≤ p,q,r ≤ ∞ satisfy

1

q+1 =

1

p+

1

r. (1.2.13)

Then for all f in Lp(G) and all g in Lr(G) satisfying∥∥g

∥∥Lr(G)

=∥∥g

∥∥Lr(G)

we have

f ∗g exists λ -a.e. and satisfies

∥∥ f ∗g∥∥

Lq(G)≤

∥∥g∥∥

Lr(G)

∥∥ f∥∥

Lp(G). (1.2.14)

Proof. Young’s inequality is proved in a way similar to Minkowski’s inequality. We

do a suitable splitting of the product | f (y)||g(y−1x)| and apply Holder’s inequality.

Observe that when r < ∞, the hypotheses on the indices imply that

1

r′+

1

q+

1

p′= 1 ,

p

q+

p

r′= 1 ,

r

q+

r

p′= 1 .

Using Holder’s inequality with exponents r′, q, and p′, we obtain

|(| f | ∗ |g|)(x)| ≤∫

G| f (y)| |g(y−1x)|dλ (y)

=∫

G| f (y)|

p

r′(| f (y)|

pq |g(y−1x)|

rq)|g(y−1x)|

rp′ dλ (y)

≤∥∥ f

∥∥p

r′Lp

(∫

G| f (y)|p|g(y−1x)|r dλ (y)

)1q(∫

G|g(y−1x)|r dλ (y)

) 1p′

=∥∥ f

∥∥p

r′Lp

(∫


)1q(∫

G|g(x−1y)|r dλ (y)

) 1p′

=

(∫


) 1q ∥∥ f

∥∥p

r′Lp

∥∥g∥∥

rp′Lr ,

where we used left invariance. Now take Lq norms (in x) and apply Fubini’s theorem

to deduce that

∥∥| f | ∗ |g|∥∥

Lq ≤∥∥ f

∥∥p

r′Lp

∥∥g∥∥

rp′Lr

(∫

G

∫

G| f (y)|p|g(y−1x)|r dλ (x)dλ (y)

)1q

=∥∥ f

∥∥p

r′Lp

∥∥g∥∥

rp′Lr

∥∥ f∥∥

pq

Lp

∥∥g∥∥ r

q

Lr

=∥∥g

∥∥Lr

∥∥ f∥∥

Lp< ∞ ,

using the hypothesis on g. This implies that | f | ∗ |g| is finite λ -a.e. and satisfies

(1.2.14); then f ∗g exists λ -a.e. and also satisfies (1.2.14).

Finally, note that if r =∞, the assumptions on p and q imply that p= 1 and q=∞,

in which case the required inequality trivially holds.

We now give a version of Theorem 1.2.12 for weak Lp spaces. Theorem 1.2.13 is

improved in Section 1.4.

Theorem 1.2.13. (Young’s inequality for weak type spaces) Let G be a locally com-

pact group with left Haar measure λ that satisfies (1.2.12). Let 1 ≤ p < ∞ and

1 < q,r < ∞ satisfy1

q+1 =

1

p+

1

r. (1.2.15)

Then there exists a constant Cp,q,r > 0 such that for all f in Lp(G) and g in Lr,∞(G),the convolution f ∗g exists λ -a.e. and satisfies

∥∥ f ∗g∥∥

Lq,∞(G)≤Cp,q,r

∥∥g∥∥

Lr,∞(G)

∥∥ f∥∥

Lp(G). (1.2.16)

Proof. As in the proofs of Theorems 1.2.10 and 1.2.12, we first obtain (1.2.16) for the

convolution of the absolute values of the functions. This implies that | f | ∗ |g| < ∞λ -a.e., and thus f ∗g exists λ -a.e. and satisfies | f ∗g| ≤ | f | ∗ |g|. We may therefore

assume that f ,g≥ 0 λ -a.e. The proof is based on a suitable splitting of the function

g. Let M be a positive real number to be chosen later. Define g1 = gχ|g|≤M and

g2 = gχ|g|>M . In view of Exercise 1.1.10 (a) we have

dg1(α) =

0 if α ≥M,

dg(α)−dg(M) if α < M,(1.2.17)

dg2(α) =

dg(α) if α > M,

dg(M) if α ≤M.(1.2.18)

Proposition 1.1.3 gives for all β > 0

d f∗g(β )≤ d f∗g1(β/2)+d f∗g2

(β/2) , (1.2.19)

and thus it suffices to estimate the distribution functions of f ∗g1 and f ∗g2. Since

g1 is the “small” part of g, it is in Ls for any s > r. In fact, we have

∫

Gg1(x)

s dλ (x) = s

∫ ∞

0αs−1dg1

(α)dα

= s

∫ M

0αs−1(dg(α)−dg(M))dα

≤ s

∫ M

0αs−1−r

∥∥g∥∥r

Lr,∞ dα− s

∫ M

0αs−1dg(M)dα

=s

s− rMs−r

∥∥g∥∥r

Lr,∞ −Msdg(M) ,

(1.2.20)

when s < ∞.

Similarly, since g2 is the “large” part of g, it is in Lt for any t < r, and

∫

Gg2(x)

t dλ (x) = t

∫ ∞

0α t−1dg2

(α)dα

= t

∫ M

0α t−1dg(M)dα+ t

∫ ∞

Mα t−1dg(α)dα

≤Mtdg(M)+ t

∫ ∞

Mα t−1−r

∥∥g∥∥r

Lr,∞ dα

≤Mt−r∥∥g

∥∥r

Lr,∞ +t

r− tMt−r

∥∥g∥∥r

Lr,∞

=r

r− tMt−r

∥∥g∥∥r

Lr,∞ . (1.2.21)

Since 1/r = 1/p′ + 1/q, it follows that 1 < r < p′. Select t = 1 and s = p′.Holder’s inequality and (1.2.20) give when p′ < ∞

|( f ∗g1)(x)| ≤∥∥ f

∥∥Lp

∥∥g1

∥∥Lp′ ≤

∥∥ f∥∥

Lp

(p′

p′− rMp′−r

∥∥g∥∥r

Lr,∞

) 1p′

(1.2.22)

and

|( f ∗g1)(x)| ≤∥∥ f

∥∥Lp M (1.2.23)

when p′ =∞. If p′ <∞ choose an M such that the right-hand side of (1.2.22) is equal

to β/2. If p′ = ∞ choose M such that the right-hand side of (1.2.23) is also equal to

β/2. That is, choose

M = (β p′2−p′rq−1‖ f‖−p′Lp ‖g‖−r

Lr,∞)1/(p′−r)

if p′ < ∞ and M = β/(2‖ f‖L1) if p′ = ∞. For these choices of M we have that

d f∗g1(β/2) = 0.

Next by Theorem 1.2.10 and (1.2.21) with t = 1 we obtain

∥∥ f ∗g2

∥∥Lp ≤

∥∥ f∥∥

Lp

∥∥g2

∥∥L1 ≤

∥∥ f∥∥

Lp

r

r−1M1−r

∥∥g∥∥r

Lr,∞ . (1.2.24)

For the value of M chosen, using (1.2.24) and Chebyshev’s inequality, we obtain

d f∗g(β ) ≤ d f∗g2(β/2)

≤ (2∥∥ f ∗g2

∥∥Lpβ

−1)p

≤ (2r∥∥ f

∥∥Lp M1−r

∥∥g∥∥r

Lr,∞(r−1)−1β−1)p

=Cqp,q,rβ

−q∥∥ f

∥∥q

Lp

∥∥g∥∥q

Lr,∞ ,

(1.2.25)

which is the required inequality. This proof gives that the constant Cp,q,r blows up

like (r−1)−p/q as r→ 1.

Example 1.2.14. Theorem 1.2.13 may fail at some endpoints:

(1) r = 1 and 1≤ p = q≤∞. On R take g(x) = 1/|x| and f = χ[0,1]. Clearly, g is in

L1,∞ and f in Lp for all 1≤ p≤∞, but the convolution of f and g is identically

equal to infinity on the interval [0,1]. Therefore, (1.2.16) fails in this case.

(2) q=∞ and 1< r = p′ <∞. On R let f (x) = (|x|1/p log |x|)−1 for |x| ≥ 2 and zero

otherwise, and also let g(x) = |x|−1/r. We see that ( f ∗ g)(x) = ∞ for |x| ≤ 1.

Thus (1.2.16) fails in this case also.

(3) r = q = ∞ and p = 1. Then inequality (1.2.16) trivially holds.

1.2.4 Approximate Identities

We now introduce the notion of approximate identities. The Banach algebra L1(G)may not have a unit element, that is, an element f0 such that

f0 ∗ f = f = f ∗ f0 (1.2.26)


for all f ∈ L1(G). In particular, this is the case when G = R; in fact, the only f0 that

satisfies (1.2.26) for all f ∈ L1(R) is not a function but the Dirac delta distribution,

introduced in Chapter 2. It is reasonable therefore to introduce the notion of approx-

imate unit or identity, a family of functions kε with the property kε ∗ f → f in L1 as

ε → 0.

Definition 1.2.15. An approximate identity (as ε→ 0) is a family of L1(G) functions

kε with the following three properties:

(i) There exists a constant c > 0 such that ‖kε‖L1(G) ≤ c for all ε > 0.

(ii)∫

G kε(x)dλ (x) = 1 for all ε > 0.

(iii) For any neighborhood V of the identity element e of the group G we have∫V c |kε(x)|dλ (x)→ 0 as ε → 0.

The construction of approximate identities on general locally compact groups G

is beyond the scope of this book and is omitted; see [152] for details. In this book we

are interested only in groups with Euclidean structure, where approximate identities

exist in abundance.

Sometimes we think of approximate identities as sequences knn. In this case

property (iii) holds as n → ∞. It is best to visualize approximate identities as se-

quences of positive functions kn that spike near 0 in such a way that the signed area

under the graph of each function remains constant (equal to one) but the support

shrinks to zero. See Figure 1.2.

Example 1.2.16. On R let P(x) = (π(x2+1))−1 and Pε(x) = ε−1P(ε−1x) for ε > 0.

Since Pε and P have the same L1 norm and

∫ +∞

−∞

1

x2 +1dx = lim

x→+∞

[arctan(x)− arctan(−x)

]= (π/2)− (−π/2) = π ,

property (ii) is satisfied. Property (iii) follows from the fact that

1

π

∫

|x|≥δ

1

ε

1

(x/ε)2 +1dx = 1− 2

πarctan(δ/ε)→ 0 as ε → 0,

for all δ > 0. The function Pε is called the Poisson kernel.

The Poisson kernel may be replaced by any integrable function of integral 1 as

the following example indicates.

Example 1.2.17. On Rn let k(x) be an integrable function with integral one. Let

kε(x) = ε−nk(ε−1x). It is straightforward to see that kε(x) is an approximate identity.

Property (iii) follows from the fact that

∫

|x|≥δ/ε|k(x)|dx→ 0

as ε → 0 for δ fixed.

-0.4 -0.2 0.2 0.4

1

2

3

4

5

6

Fig. 1.2 The Fejer kernel F5 plotted on the interval [− 12, 1

2].

Example 1.2.18. On the circle group T1 let

FN(t) =N

∑j=−N

(1− | j|

N +1

)e2πi jt =

1

N +1

(sin(π(N +1)t)

sin(πt)

)2

. (1.2.27)

To check the previous equality we use that

sin2(x) = (2− e2ix− e−2ix)/4 ,

and we carry out the calculation. FN is called the Fejer kernel. See Figure 1.2. To

see that the sequence FNN is an approximate identity, we check conditions (i), (ii),

and (iii) in Definition 1.2.15. Property (iii) follows from the expression giving FN in

terms of sines, while property (i) follows from the expression giving FN in terms of

exponentials. Property (ii) is identical to property (i), since FN is nonnegative.

Next comes the basic theorem concerning approximate identities.

Theorem 1.2.19. Let kε be an approximate identity on a locally compact group G

with left Haar measure λ .

(1) If f lies in Lp(G) for 1≤ p < ∞, then ‖kε ∗ f − f‖Lp(G)→ 0 as ε → 0.

(2) Let f be a function in L∞(G) that is uniformly continuous on a subset K of

G, in the sense that for all δ > 0 there is a neighborhood V of the identity

element such that for all x ∈ K and y ∈V we have | f (y−1x)− f (x)|< δ . Then

we have that ‖kε ∗ f − f‖L∞(K)→ 0 as ε→ 0. In particular, if f is bounded and

continuous at a point x0 ∈ G, then (kε ∗ f )(x0)→ f (x0) as ε → 0.

Proof. We start with the case 1 ≤ p < ∞. We recall that continuous functions with

compact support are dense in Lp of locally compact Hausdorff spaces equipped

with measures arising from nonnegative linear functionals; see [152, Theorem

12.10]. For a continuous function g supported in a compact set L we have we have

|g(h−1x)− g(x)|p ≤ (2‖g‖L∞)pχW−1L for h in a relatively compact neighborhood

W of the identity element e. By the Lebesgue dominated convergence theorem we

obtain ∫

G|g(h−1x)−g(x)|p dλ (x)→ 0 (1.2.28)

as h→ e. Now approximate a given f in Lp(G) by a continuous function with com-

pact support g to deduce that

∫

G| f (h−1x)− f (x)|p dλ (x)→ 0 as h→ e . (1.2.29)

Because of (1.2.29), given a δ > 0 there exists a neighborhood V of e such that

h ∈V =⇒∫

G| f (h−1x)− f (x)|p dλ (x)<

(δ

2c

)p

, (1.2.30)

where c is the constant that appears in Definition 1.2.15 (i). Since kε has integral one

for all ε > 0, we have

(kε ∗ f )(x)− f (x) = (kε ∗ f )(x)− f (x)∫

Gkε(y)dλ (y)

=

∫

G( f (y−1x)− f (x))kε(y)dλ (y)

=∫

V( f (y−1x)− f (x))kε(y)dλ (y)

+

∫

V c( f (y−1x)− f (x))kε(y)dλ (y) .

(1.2.31)

Now take Lp norms in x in (1.2.31). In view of (1.2.30),

∥∥∥∥∫

V( f (y−1x)− f (x))kε(y)dλ (y)

∥∥∥∥Lp(G,dλ (x))

≤∫

V

∥∥ f (y−1x)− f (x)∥∥

Lp(G,dλ (x))|kε(y)|dλ (y)

≤∫

V

δ

2c|kε(y)|dλ (y)<

δ

2,

(1.2.32)

while∥∥∥∥∫

V c( f (y−1x)− f (x))kε(y)dλ (y)

∥∥∥∥Lp(G,dλ (x))

≤∫

V c2∥∥ f

∥∥Lp(G)

|kε(y)|dλ (y)<δ

2,

(1.2.33)

provided we have that

∫

V c|kε(x)|dλ (x)<

δ

4(∥∥ f

∥∥Lp +1

) . (1.2.34)

Choose ε0 > 0 such that (1.2.34) is valid for ε < ε0 by property (iii). Now (1.2.32)

and (1.2.33) imply the required conclusion.

The case p = ∞ follows similarly. Let f be a bounded function on G that is

uniformly continuous on K. Given δ > 0, there is a neighborhood V of e such that,

whenever y ∈V and x ∈ K we have

| f (y−1x)− f (x)|< δ

2c, (1.2.35)

where c is as in Definition 1.2.15 (i). By property (iii) in Definition 1.2.15, there is

an ε0 > 0 such that for 0 < ε < ε0 we have

∫

V c|kε(y)|dλ (y)<

δ

4(‖ f‖L∞(G)+1

) . (1.2.36)

Using (1.2.35) and (1.2.36), we deduce that

supx∈K

|(kε ∗ f )(x)− f (x)|

≤∫

V|kε(y)|sup

x∈K

| f (y−1x)− f (x)|dλ (y)+∫

V c|kε(y)|sup

x∈K

| f (y−1x)− f (x)|dλ (y)

≤ cδ

2c+

δ

4(‖ f‖L∞(G)+1

) 2‖ f‖L∞(G) ≤ δ .

This shows that kε ∗ f converge uniformly to f on K as ε → 0. In particular, if

K = x0 and f is bounded and continuous at x0, we have (kε ∗ f )(x0)→ f (x0).

Remark 1.2.20. Observe that if Haar measure satisfies (1.2.12), then the conclusion

of Theorem 1.2.19 also holds for f ∗ kε .

A simple modification in the proof of Theorem 1.2.19 yields the following vari-

ant, which presents a significant difference only when a = 0.

Theorem 1.2.21. Let kε be a family of functions on a locally compact group G that

satisfies properties (i) and (iii) of Definition 1.2.15 and also

∫

Gkε(x)dλ (x) = a

for some fixed a ∈ C and for all ε > 0. Let f ∈ Lp(G) for some 1≤ p≤ ∞ .

(a) If 1≤ p < ∞, then ‖kε ∗ f −a f‖Lp(G)→ 0 as ε → 0 .

(b) If p = ∞ and f is uniformly continuous on a subset K of G, in the sense that

for any δ > 0 there is a neighborhood V of the identity element of G such that

supx∈G supy∈V | f (y−1x)− f (x)| ≤ δ , then we have that ‖kε ∗ f −a f‖L∞(K)→ 0

as ε → 0.

Exercises

1.2.1. Let G be a locally compact group and let f ,g in L1(G) be supported in the

subsets A and B of G, respectively. Prove that f ∗ g is supported in the algebraic

product set AB.

1.2.2. For a function f on a locally compact group G and t ∈ G, let t f (x) = f (tx)and f t(x) = f (xt). Show that

t f ∗g = t( f ∗g) and f ∗gt = ( f ∗g)t

whenever f ,g ∈ L1(G), equipped with left Haar measure.

1.2.3. Let G be a locally compact group with left Haar measure. Let f ∈ Lp(G)

and g ∈ Lp′(G), where 1 0 there exists a relatively compact symmet-

ric neighborhood of the origin U such that u ∈U implies ‖ug− g‖Lp′ (G)

< ε and

therefore

|( f ∗g)(v)− ( f ∗g)(w)|<∥∥ f

∥∥Lp ε

whenever v−1w ∈U .

1.2.4. (a) Prove that compactly supported functions are dense in Lp(Rn) for all 0 <p < ∞.

(b) Show that smooth functions with compact support are dense in Lp(Rn) for all

1≤ p < ∞.[Hint: Part (b): Use Theorem 1.2.19 with kε(x) = ε−nk(ε−1x) and k smooth and

compactly supported function.]

1.2.5. Show that a Haar measure λ for the multiplicative group of all positive real

numbers is

λ (A) =∫ ∞

0χA(t)

dt

t.

1.2.6. Let G = R2 \(0,y) : y∈R with group operation (x,y)(z,w) = (xz,xw+y).[Think of G as the group of all 2× 2 matrices with bottom row (0,1) and nonzero

top left entry.] Show that a left Haar measure on G is

λ (A) =∫ +∞

−∞

∫ +∞

−∞χA(x,y)

dxdy

x2,

while a right Haar measure on G is

ρ(A) =∫ +∞

−∞

∫ +∞

−∞χA(x,y)

dxdy

|x| .

1.2.7. ([144], [145]) Use Theorem 1.2.10 to prove that

(∫ ∞

0

(1

x

∫ x

0| f (t)|dt

)p

dx

)1p

≤ p

p−1

∥∥ f∥∥

Lp(0,∞),

(∫ ∞

0

(∫ ∞

x| f (t)|dt

)p

dx

)1p

≤ p

(∫ ∞

0| f (t)|pt p dt

)1p

,

when 1 < p < ∞.[Hint: On the multiplicative group (R+, dt

t) consider the convolution of the function

| f (x)|x1p with the function x

− 1p′ χ[1,∞) and the convolution of the function | f (x)|x1+ 1

p

with x1p χ(0,1].

]

1.2.8. (G. H. Hardy) Let 0 < b < ∞ and 1≤ p < ∞. Prove that

(∫ ∞

0

(∫ x

0| f (t)|dt

)p

x−b−1 dx

)1p

≤ p

b

(∫ ∞

0| f (t)|pt p−b−1 dt

)1p

,

(∫ ∞

0

(∫ ∞

x| f (t)|dt

)p

xb−1 dx

)1p

≤ p

b

(∫ ∞

0| f (t)|pt p+b−1 dt

)1p

.

[Hint: On the multiplicative group (R+, dt

t) consider the convolution of the function

| f (x)|x1− bp with x

− bp χ[1,∞) and of the function | f (x)|x1+ b

p with xbp χ(0,1].

]

1.2.9. On Rn let T ( f ) = f ∗K, where K is a positive L1 function and f is in Lp,

1≤ p≤ ∞. Prove that the operator norm of T : Lp → Lp is equal to ‖K‖L1 .[Hint: Clearly, ‖T‖Lp→Lp ≤ ‖K‖L1 . Conversely, fix 0 < ε < 1 and let N be a positive

integer. Let χN = χB(0,N) and for any R > 0 let KR = KχB(0,R), where B(x,R) is the

ball of radius R centered at x. Observe that for |x| ≤ (1− ε)N, we have B(0,Nε) B(x,N); thus

∫Rn χN(x− y)KNε(y)dy =

∫Rn KNε(y)dy =

∥∥KNε

∥∥L1 . Then

∥∥K ∗χN

∥∥p

Lp

‖χN‖pLp

≥∥∥KNε ∗χN

∥∥p

Lp(B(0,(1−ε)N)∥∥χN

∥∥p

Lp

≥∥∥KNε

∥∥p

L1(1− ε)n .

Let N → ∞ first and then ε → 0.]

1.2.10. On the multiplicative group (R+, dtt) let T ( f ) = f ∗K, where K is a positive

L1 function and f is in Lp, 1≤ p≤ ∞. Prove that the operator norm of T : Lp → Lp

is equal to the L1 norm of K. Deduce that the constants p/(p−1) and p/b are sharp

in Exercises 1.2.7 and 1.2.8.[Hint: Adapt the idea of Exercise 1.2.9 to this setting.

]

1.2.11. Let Qk(t) = ck(1− t2)k for t ∈ [−1,1] and zero elsewhere, where ck is cho-

sen such that∫ 1−1 Qk(t)dt = 1 for all k = 1,2, . . . .

(a) Show that ck <√

k.


(b) Use part (a) to show that Qkk is an approximate identity on R as k→ ∞.

(c) Given a continuous function f on R that vanishes outside the interval [−1,1],show that f ∗Qk converges to f uniformly on [−1,1] as k→ ∞.

(d) (Weierstrass) Prove that every continuous function on [−1,1] can be approxi-

mated uniformly by polynomials.[Hint: Part (a): Estimate the integral

∫|t|≤k−1/2 Qk(t)dt from below using the in-

equality (1− t2)k ≥ 1− kt2 for |t| ≤ 1. Part (d): Consider the function g(t) =f (t)− f (−1)− t+1

2( f (1)− f (−1)).

]

1.2.12. Show that the Laplace transform L( f )(x) =∫ ∞

0 f (t)e−xtdt maps L2(0,∞) to

itself with norm at most√π .[

Hint: Consider convolution with the kernel√

t e−t on the group L2((0,∞), dtt).]

1.2.13. ([62]) Let F ≥ 0, G≥ 0 be measurable functions on the sphere Sn−1 and let

K ≥ 0 be a measurable function on [−1,1]. Prove that

∫

Sn−1

∫

Sn−1F(θ)G(ϕ)K(θ ·ϕ)dϕ dθ ≤C‖F‖Lp(Sn−1)‖G‖Lp′ (Sn−1)

,

where 1≤ p≤∞, θ ·ϕ =∑nj=1 θ jϕ j and C =

∫Sn−1 K(θ ·ϕ)dϕ, which is independent

of θ . Moreover, show that C is the best possible constant in the preceding inequality.

Using duality, compute the norm of the linear operator

F(θ) →∫

Sn−1F(θ)K(θ ·ϕ)dϕ

from Lp(Sn−1) to itself.[Hint: Observe that

∫Sn−1

∫Sn−1 F(θ)G(ϕ)K(θ ·ϕ)dϕ dθ is bounded by the quantity

∫

Sn−1

[∫

Sn−1F(θ)K(θ ·ϕ)dθ

]p

dϕ

1p

‖G‖Lp′ (Sn−1)

.

Apply Holder’s inequality to the functions F and 1 with respect to the measure

K(θ ·ϕ)dθ to deduce that∫

Sn−1 F(θ)K(θ ·ϕ)dθ is controlled by

(∫

Sn−1F(θ)pK(θ ·ϕ)dθ

)1/p(∫

Sn−1K(θ ·ϕ)dθ

)1/p′

.

Use Fubini’s theorem to bound the latter by

‖F‖Lp(Sn−1)‖G‖Lp′ (Sn−1)

∫

Sn−1K(θ ·ϕ)dϕ.

Note that equality is attained if and only if both F and G are constants.]

1.3 Interpolation 33

1.3 Interpolation

The theory of interpolation of operators is vast and extensive. In this section we

are mainly concerned with a couple of basic interpolation results that appear in a

variety of applications and constitute the foundation of the field. These results are

the Marcinkiewicz interpolation theorem and the Riesz–Thorin interpolation theo-

rem. These theorems are traditionally proved using real and complex variables tech-

niques, respectively. A byproduct of the Riesz–Thorin interpolation theorem, Stein’s

theorem on interpolation of analytic families of operators, has also proved to be an

important and useful tool in many applications and is presented at the end of the

section.

We begin by setting up the background required to formulate the results of this

section. Let (X ,µ) and (Y,ν) be two measure spaces. Suppose we are given a linear

operator T , initially defined on the set of simple functions on X , such that for all f

simple on X , T ( f ) is a ν-measurable function on Y . Let 0 0 such that for all simple functions f on X we have

∥∥T ( f )∥∥

Lq(Y,ν)≤Cp,q

∥∥ f∥∥

Lp(X ,µ), (1.3.1)

then by density, T admits a unique bounded extension from Lp(X ,µ) to Lq(Y,ν).This extension is also denoted by T . Operators that map Lp to Lq are called of strong

type (p,q) and operators that map Lp to Lq,∞ are called weak type (p,q).

1.3.1 Real Method: The Marcinkiewicz Interpolation Theorem

Definition 1.3.1. Let T be an operator defined on a linear space of complex-valued

measurable functions on a measure space (X ,µ) and taking values in the set of all

complex-valued finite almost everywhere measurable functions on a measure space

(Y,ν). Then T is called linear if for all f , g in the domain of T and all λ ∈ C we

have

T ( f +g) = T ( f )+T (g) and T (λ f ) = λT ( f ). (1.3.2)

T is called sublinear if for all f , g in the domain of T and all λ ∈ C we have

|T ( f +g)| ≤ |T ( f )|+ |T (g)| and |T (λ f )|= |λ ||T ( f )|. (1.3.3)

T is called quasi-linear if for all f , g in the domain of T and all λ ∈ C we have

|T ( f +g)| ≤ K(|T ( f )|+ |T (g)|) and |T (λ f )|= |λ ||T ( f )| (1.3.4)

for some constant K > 0. Sublinearity is a special case of quasi-linearity.

For instance, T1 and T2 are linear operators, then (|T1|p + |T2|p)1/p is sublinear if

p≥ 1 and quasi-linear if 0 < p < 1.

Theorem 1.3.2. Let (X ,µ) be a σ -finite measure space, let (Y,ν) be another mea-

sure space, and let 0 < p0 < p1 ≤ ∞. Let T be a sublinear operator defined on

Lp0(X)+Lp1(X) = f0 + f1 : f j ∈ Lp j(X j), j = 0,1 and taking values in the space

of measurable functions on Y . Assume that there exist A0,A1 < ∞ such that

∥∥T ( f )∥∥

Lp0 ,∞(Y )≤ A0

∥∥ f∥∥

Lp0 (X)for all f ∈ Lp0(X) , (1.3.5)

∥∥T ( f )∥∥

Lp1 ,∞(Y )≤ A1

∥∥ f∥∥

Lp1 (X)for all f ∈ Lp1(X) . (1.3.6)

Then for all p0 < p < p1 and for all f in Lp(X) we have the estimate

∥∥T ( f )∥∥

Lp(Y )≤ A

∥∥ f∥∥

Lp(X), (1.3.7)

where

A = 2

(p

p− p0+

p

p1− p

)1p

A

1p− 1

p11

p0− 1

p10 A

1p0− 1

p

1p0− 1

p11 . (1.3.8)

Proof. Assume first that p1 < ∞. Fix f a function in Lp(X) and α > 0. We split

f = f α0 + f α1 , where f α0 is in Lp0 and f α1 is in Lp1 . The splitting is obtained by

cutting | f | at height δα for some δ > 0 to be determined later. Set

f α0 (x) =

f (x) for | f (x)|> δα,

0 for | f (x)| ≤ δα,

f α1 (x) =

f (x) for | f (x)| ≤ δα,

0 for | f (x)|> δα.

It can be checked easily that f α0 (the unbounded part of f ) is an Lp0 function and

that f α1 (the bounded part of f ) is an Lp1 function. Indeed, since p0 < p, we have

∥∥ f α0∥∥p0

Lp0=

∫

| f |>δα| f (x)|p| f (x)|p0−p dµ(x)≤ (δα)p0−p

∥∥ f∥∥p

Lp

and similarly, since p < p1,

∥∥ f α1∥∥p1

Lp1≤ (δα)p1−p

∥∥ f∥∥p

Lp .

In view of the subadditivity property of T contained in (1.3.3) we obtain that

|T ( f )| ≤ |T ( f α0 )|+ |T ( f α1 )| ,

which implies

y∈Y : |T ( f )(y)|>α y∈Y : |T ( f α0 )(y)|>α/2∪y∈Y : |T ( f α1 )(y)|>α/2,

and therefore

dT ( f )(α)≤ dT ( fα0 )(α/2)+dT ( fα1 )(α/2) . (1.3.9)

Hypotheses (1.3.5) and (1.3.6) together with (1.3.9) now give

dT ( f )(α)≤A

p00

(α/2)p0

∫

| f |>δα| f (x)|p0 dµ(x)+

Ap11

(α/2)p1

∫

| f |≤δα| f (x)|p1 dµ(x).

In view of the last estimate and Proposition 1.1.4, we obtain that

∥∥T ( f )∥∥p

Lp ≤ p(2A0)p0

∫ ∞

0α p−1α−p0

∫

| f |>δα| f (x)|p0 dµ(x)dα

+ p(2A1)p1

∫ ∞

0α p−1α−p1

∫

| f |≤δα| f (x)|p1 dµ(x)dα

= p(2A0)p0

∫

X| f (x)|p0

∫ 1δ| f (x)|

0α p−1−p0 dα dµ(x)

+ p(2A1)p1

∫

X| f (x)|p1

∫ ∞

1δ| f (x)|

α p−1−p1 dα dµ(x)

=p(2A0)

p0

p− p0

1

δ p−p0

∫

X| f (x)|p0 | f (x)|p−p0 dµ(x)

+p(2A1)

p1

p1− p

1

δ p−p1

∫

X| f (x)|p1 | f (x)|p−p1 dµ(x)

= p

((2A0)

p0

p− p0

1

δ p−p0+

(2A1)p1

p1− pδ p1−p

)∥∥ f∥∥p

Lp ,

and the convergence of the integrals in α is justified from p0 0 such that

(2A0)p0

1

δ p−p0= (2A1)

p1δ p1−p ,

and observe that the last displayed constant is equal to the pth power of the constant

in (1.3.8). We have therefore proved the theorem when p1 < ∞.

We now consider the case p1 = ∞. Write f = f α0 + f α1 , where

f α0 (x) =

f (x) for | f (x)|> γα,

0 for | f (x)| ≤ γα,

f α1 (x) =

f (x) for | f (x)| ≤ γα,

0 for | f (x)|> γα.

We have ∥∥T ( f α1 )∥∥

L∞≤ A1

∥∥ f α1∥∥

L∞≤ A1γα = α/2 ,


provided we choose γ = (2A1)−1. It follows that the set y ∈Y : |T ( f α1 )(y)|> α/2

has measure zero. Therefore,

dT ( f )(α)≤ dT ( fα0 )(α/2).

Since T maps Lp0 to Lp0,∞ with norm at most A0, it follows that

dT ( fα0 )(α/2)≤(2A0)

p0∥∥ f α0

∥∥p0

Lp0

α p0=

(2A0)p0

α p0

∫

| f |>γα| f (x)|p0 dµ(x). (1.3.10)

Using (1.3.10) and Proposition 1.1.4, we obtain

∥∥T ( f )∥∥p

Lp = p

∫ ∞

0α p−1dT ( f )(α)dα

≤ p

∫ ∞

0α p−1dT ( fα0 )(α/2)dα

≤ p

∫ ∞

0α p−1 (2A0)

p0

α p0

∫

| f |>α/(2A1)| f (x)|p0 dµ(x)dα

= p(2A0)p0

∫

X| f (x)|p0

∫ 2A1| f (x)|

0α p−p0−1 dα dµ(x)

=p(2A1)

p−p0(2A0)p0

p− p0

∫

X| f (x)|p dµ(x) .

This proves the theorem with constant

A = 2

(p

p− p0

) 1p

A1− p0

p

1 A

p0p

0 . (1.3.11)

Observe that when p1 = ∞, the constant in (1.3.11) coincides with that in (1.3.8).

Remark 1.3.3. Notice that the proof of Theorem 1.3.2 only makes use of the subad-

ditivity property |T ( f+g)| ≤ |T ( f )|+|T (g)| of T in hypothesis (1.3.3).

If T is a linear operator (instead of sublinear), then we can relax the hypotheses

of Theorem 1.3.2 by assuming that (1.3.5) and (1.3.6) hold for all simple functions

f on X . Then the functions f α0 and f α1 constructed in the proof are also simple, and

we conclude that (1.3.7) holds for all simple functions f on X . By density, T has a

unique extension on Lp(X) that also satisfies (1.3.7).

1.3.2 Complex Method: The Riesz–Thorin Interpolation Theorem

The next interpolation theorem assumes stronger endpoint estimates, but yields a

more natural bound on the norm of the operator on the intermediate spaces. Unfor-

tunately, it is mostly applicable for linear operators and in some cases for sublinear

operators (often via a linearization process) but it does not apply to quasi-linear

operators without some loss in the constant.

Recall that a simple function is called finitely simple if it is supported in a set

of finite measure. Finitely simple functions are dense in Lp(X ,µ) for 0 < p < ∞,

whenever (X ,µ) is a σ -finite measure space.

Theorem 1.3.4. Let (X ,µ) and (Y,ν) be two σ -finite measure spaces. Let T be a

linear operator defined on the set of all finitely simple functions on X and taking

values in the set of measurable functions on Y . Let 1≤ p0, p1,q0,q1 ≤∞ and assume

that∥∥T ( f )

∥∥Lq0

≤M0

∥∥ f∥∥

Lp0,

∥∥T ( f )∥∥

Lq1≤M1

∥∥ f∥∥

Lp1,

(1.3.12)

for all finitely simple functions f on X. Then for all 0 < θ < 1 we have

∥∥T ( f )∥∥

Lq ≤M1−θ0 Mθ

1

∥∥ f∥∥

Lp (1.3.13)

for all finitely simple functions f on X, where

1

p=

1−θp0

+θ

p1and

1

q=

1−θq0

+θ

q1. (1.3.14)

Consequently, when p < ∞, by density, T has a unique bounded extension from

Lp(X ,µ) to Lq(Y,ν) when p and q are as in (1.3.14).

Proof. Let

f =m

∑k=1

akeiαkχAk

be a finitely simple function on X , where ak > 0, αk are real, and Ak are pairwise

disjoint subsets of X with finite measure.

We need to control

∥∥T ( f )∥∥

Lq(Y,ν)= sup

g

∣∣∣∣∫

YT ( f )(y)g(y)dν(y)

∣∣∣∣ ,

where the supremum is taken over all finitely simple functions g on Y with Lq′ norm

less than or equal to 1. Write

g =n

∑j=1

b jeiβ jχB j

,

where b j > 0, β j are real, and B j are pairwise disjoint subsets of Y with finite ν-

measure. Let

P(z) =p

p0(1− z)+

p

p1z and Q(z) =

q′

q′0(1− z)+

q′

q′1z . (1.3.15)


For z in the closed strip S = z ∈ C : 0≤ Rez≤ 1, define

fz =m

∑k=1

aP(z)k eiαkχAk

, gz =n

∑j=1

bQ(z)j eiβ jχB j

, (1.3.16)

and

F(z) =∫

YT ( fz)(y)gz(y)dν(y) .

Notice that fθ = f and gθ = f . By linearity we have

F(z) =m

∑k=1

n

∑j=1

aP(z)k b

Q(z)j eiαk eiβ j

∫

YT (χAk

)(y)χB j(y)dν(y) .

Since ak,b j > 0, F is analytic in z, and the expression

∫

YT (χAk

)(y)χB j(y)dν(y)

is a finite constant, being an absolutely convergent integral; this is seen by Holder’s

inequality with exponents q0 and q′0 (or q1 and q′1) and (1.3.12).

By the disjointness of the sets Ak we have (even when p0 = ∞)

∥∥ fit

∥∥Lp0

=∥∥ f

∥∥p

p0Lp ,

since |aP(it)k |= a

pp0k , and by the disjointness of the B j’s we have (even when q0 = 1)

∥∥git

∥∥L

q′0=

∥∥g∥∥

q′q′0

Lq′ ,

since |bQ(it)j |= b

q′q′0

j . Thus Holder’s inequality and the hypothesis give

|F(it)| ≤∥∥T ( fit)

∥∥Lq0

∥∥git

∥∥L

q′0

≤M0

∥∥ fit

∥∥Lp0

∥∥git

∥∥L

q′0

= M0

∥∥ f∥∥

pp0Lp

∥∥g∥∥

q′q′0

Lq′ .

(1.3.17)

By similar calculations, which are valid even when p1 = ∞ and q1 = 1, we have

∥∥ f1+it

∥∥Lp1

=∥∥ f

∥∥p

p1Lp

and

∥∥g1+it

∥∥L

q′1=

∥∥g∥∥

q′q′1

Lq′ .

Also, in a way analogous to that we obtained (1.3.17) we deduce that

|F(1+ it)| ≤M1

∥∥ f∥∥

pp1Lp

∥∥g∥∥

q′q′1

Lq′ . (1.3.18)

To finish the proof we will need the following lemma, known as Hadamard’s

three lines lemma.

Lemma 1.3.5. Let F be analytic in the open strip S = z ∈ C : 0 < Re z < 1,continuous and bounded on its closure, such that |F(z)| ≤ B0 when Re z = 0 and

|F(z)| ≤ B1 when Re z = 1, for some 0 < B0,B1 < ∞. Then |F(z)| ≤ B1−θ0 Bθ1 when

Re z = θ , for any 0≤ θ ≤ 1.

To prove the lemma we define analytic functions

G(z) = F(z)(B1−z0 Bz

1)−1 and Gn(z) = G(z)e(z

2−1)/n

for z in the unit strip S, for n = 1,2, . . . . Since F is bounded on S and

|B1−z0 Bz

1| ≥min(1,B0)min(1,B1)> 0

for all z ∈ S, we conclude that G is bounded by some constant M on S. Since

|Gn(x+ iy)| ≤Me−y2/ne(x2−1)/n ≤Me−y2/n ,

we deduce that Gn(x+ iy) converges to zero uniformly in 0 ≤ x ≤ 1 as |y| → ∞.

Select y(n) > 0 such that for |y| ≥ y(n), we have |Gn(x+ iy)| ≤ 1 for all x ∈ [0,1].Also, the assumptions on F imply that G is bounded by one on the two lines forming

the boundary of S. By the maximum principle we obtain that |Gn(z)| ≤ 1 for all z in

the rectangle [0,1]× [−y(n),y(n)]; hence |Gn(z)| ≤ 1 everywhere in the closed strip.

Letting n→∞, we conclude that |G(z)| ≤ 1 in the closed strip. Taking z = θ + it we

deduce that

|F(θ + it)| ≤ |B1−θ−it0 Bθ+it

1 |= B1−θ0 Bθ1

whenever t is real. This proves the required conclusion. Returning to the proof of Theorem 1.3.4, we observe that F is analytic in the open

strip S and continuous on its closure. Also, F is bounded on the closed unit strip (by

some constant that depends on f and g). Therefore, (1.3.17), (1.3.18), and Lemma

1.3.5 give

|F(z)| ≤(

M0

∥∥ f∥∥

pp0Lp

∥∥g∥∥

q′q′0

Lq′

)1−θ(M1

∥∥ f∥∥

pp1Lp

∥∥g∥∥

q′q′1

Lq′

)θ= M1−θ

0 Mθ1

∥∥ f∥∥

Lp

∥∥g∥∥

Lq′ ,

when Rez = θ . Observe that P(θ) = Q(θ) = 1 and hence

F(θ) =∫

YT ( f )gdν .

Taking the supremum over all finitely simple functions g on Y with Lq′ norm less

than or equal to one, we conclude the proof of the theorem.

We now give an application of Theorem 1.3.4.

Example 1.3.6. One may prove Young’s inequality (Theorem 1.2.12) using the

Riesz–Thorin interpolation theorem (Theorem 1.3.4). Fix a function g in Lr and

let T ( f ) = f ∗g. Since T : L1 → Lr with norm at most ‖g‖Lr and T : Lr′ → L∞ with

norm at most ‖g‖Lr , Theorem 1.3.4 gives that T maps Lp to Lq with norm at most

the quantity ‖g‖θLr‖g‖1−θLr = ‖g‖Lr , where

1

p=

1−θ1

+θ

r′and

1

q=

1−θr

+θ

∞. (1.3.19)

Finally, observe that equations (1.3.19) give (1.2.13).

1.3.3 Interpolation of Analytic Families of Operators

Theorem 1.3.4 can be extended to the case in which the interpolated operators are

allowed to vary. In particular, if a family of operators depends analytically on a

parameter z, then the proof of this theorem can be adapted to work in this setting.

We describe the setup for this theorem. Let (X ,µ) and (Y,ν) be σ -finite measure

spaces. Suppose that for every z in the closed strip S = z ∈ C : 0≤ Rez≤ 1 there

is an associated linear operator Tz defined on the space of finitely simple functions

on X and taking values in the space of measurable functions on Y such that

∫

Y|Tz(χA)χB|dν < ∞ (1.3.20)

whenever A and B are subsets of finite measure of X and Y , respectively. The family

Tzz is said to be analytic if for all f ,g finitely simple functions we have that the

function

z →∫

YTz( f )gdν (1.3.21)

is analytic in the open strip S = z∈C : 0 < Rez < 1 and continuous on its closure.

The analytic family Tzz is called of admissible growth if there is a constant τ0 with

0≤ τ0 < π such that for finitely simple functions f on X and g on Y there is constant

C( f ,g) such that

log

∣∣∣∣∫

YTz( f )gdν

∣∣∣∣≤C( f ,g)eτ0|Imz| (1.3.22)

for all z satisfying 0 ≤ Rez ≤ 1. Note that if there is τ0 ∈ (0,π) such that for all

measurable subsets A of X and B of Y of finite measure there is a constant c(A,B)such that

log

∣∣∣∣∫

BTz(χA)dν

∣∣∣∣≤ c(A,B)eτ0|Imz| , (1.3.23)

then (1.3.22) holds for f = ∑Mk=1 akχAk

and g = ∑Nj=1 b jχB j

and

C( f ,g) = log(MN)+M

∑k=1

N

∑j=1

(c(Ak,B j)+

∣∣ log |ak b j|∣∣) .

The extension of the Riesz–Thorin interpolation theorem is as follows.

Theorem 1.3.7. Let Tz be an analytic family of linear operators of admissible growth

defined on the space of finitely simple functions of a σ -finite measure space (X ,µ)and taking values in the set of measurable functions of another σ -finite measure

space (Y,ν). Let 1 ≤ p0, p1,q0,q1 ≤ ∞ and suppose that M0 and M1 are positive

functions on the real line such that for some τ1 with 0≤ τ1 < π we have

sup−∞<y<+∞

e−τ1|y| logM j(y)< ∞ (1.3.24)

for j = 0,1. Fix 0 < θ < 1 and define p,q by the equations

1

p=

1−θp0

+θ

p1and

1

q=

1−θq0

+θ

q1. (1.3.25)

Suppose that for all finitely simple functions f on X we have

∥∥Tiy( f )∥∥

Lq0≤M0(y)

∥∥ f∥∥

Lp0, (1.3.26)

∥∥T1+iy( f )∥∥

Lq1≤M1(y)

∥∥ f∥∥

Lp1. (1.3.27)

Then for all finitely simple functions f on X we have

∥∥Tθ ( f )∥∥

Lq ≤M(θ)∥∥ f

∥∥Lp (1.3.28)

where for 0 < x < 1

M(x) = exp

sin(πx)

2

∫ ∞

−∞

[logM0(t)

cosh(πt)−cos(πx)+

logM1(t)

cosh(πt)+cos(πx)

]dt

.

Thus, by density, Tθ has a unique bounded extension from Lp(X ,µ) to Lq(Y,ν) when

p and q are as in (1.3.25).

Note that in view of (1.3.24), the integral defining M(t) converges absolutely. The

proof of the previous theorem is based on an extension of Lemma 1.3.5.

Lemma 1.3.8. Let F be analytic on the open strip S = z ∈ C : 0 < Re z < 1 and

continuous on its closure such that for some A < ∞ and 0≤ τ0 < π we have

log |F(z)| ≤ Aeτ0|Im z| (1.3.29)

for all z ∈ S. Then

|F(x+ iy)| ≤ exp

sin(πx)

2

∫ ∞

−∞

[log |F(it + iy)|

cosh(πt)− cos(πx)+

log |F(1+ it + iy)|cosh(πt)+ cos(πx)

]dt

whenever 0 < x < 1, and y is real.

Assuming Lemma 1.3.8, we prove Theorem 1.3.7.

Proof. Fix 0 < θ < 1 and finitely simple functions f on X and g on Y such that

‖ f‖Lp = ‖g‖Lq′ = 1. Note that since 0 < θ < 1 we must have 1 < p,q < ∞. Let

f =m

∑k=1

akeiαkχAkand g =

n

∑j=1

b jeiβ jχB j

,

where ak > 0, b j > 0, αk, β j are real, Ak are pairwise disjoint subsets of X with finite

measure, and B j are pairwise disjoint subsets of Y with finite measure for all k, j.

Let P(z), Q(z) be as in (1.3.15) and fz, gz as in (1.3.16). Define for z ∈ S

F(z) =∫

YTz( fz)gz dν . (1.3.30)

Linearity gives that

F(z) =m

∑k=1

n

∑j=1

aP(z)k b

Q(z)j eiαk eiβ j

∫

YTz(χAk

)(x)χB j(x)dν(x) ,

and conditions (1.3.20) together with the fact that Tzz is an analytic family imply

that F(z) is a well-defined analytic function on the unit strip that extends continu-

ously to its boundary.

Since Tzz is a family of admissible growth, (1.3.23) holds for some c(Ak,B j)and τ0 ∈ (0,π) and this combined with the facts that

|aP(z)k | ≤ a

pp0

+ pp1

k and |bQ(z)j | ≤ b

q′q′0

+ q′q′1

j

for all z with 0 < Re z < 1, implies (1.3.29) with τ0 as in (1.3.23) and

A = log(mn)+m

∑k=1

n

∑j=1

(c(Ak,B j)+

( p

p0+

p

p1

)∣∣ log ak

∣∣+( q′

q′0+

q′

q′1

)∣∣ log b j

∣∣).

Thus F satisfies the hypotheses of Lemma 1.3.8. Moreover, the calculations in the

proof of Theorem 1.3.4 show that (even when p0 = ∞, q0 = 1, p1 = ∞, q1 = 1)

∥∥ fiy

∥∥Lp0

=∥∥ f

∥∥p

p0Lp = 1 =

∥∥g∥∥

q′q′0

Lq′ =∥∥giy

∥∥L

q′0

when y ∈ R , (1.3.31)

∥∥ f1+iy

∥∥Lp1

=∥∥ f

∥∥p

p1Lp = 1 =

∥∥g∥∥

q′q′1

Lq′ =∥∥g1+iy

∥∥L

q′1

when y ∈ R . (1.3.32)

Holder’s inequality, (1.3.31), and the hypothesis (1.3.26) now give

|F(iy)| ≤∥∥Tiy( fiy)

∥∥Lq0

∥∥giy

∥∥L

q′0≤M0(y)

∥∥ fiy

∥∥Lp0

∥∥giy

∥∥L

q′0= M0(y)

for all y real. Similarly, (1.3.32), and (1.3.27) imply

|F(1+ iy)| ≤∥∥T1+iy( f1+iy)

∥∥Lq1

∥∥g1+iy

∥∥L

q′1≤M1(y)

∥∥ f1+iy

∥∥Lp1

∥∥g1+iy

∥∥L

q′1= M1(y)

for all y ∈ R. These inequalities and the conclusion of Lemma 1.3.8 yield

|F(x)| ≤ exp

sin(πx)

2

∫ ∞

−∞

[logM0(t)


logM1(t)

cosh(πt)+cos(πx)

]dt

=M(x)

for all 0 < x < 1. But notice that

F(θ) =∫

YTθ ( f )gdν . (1.3.33)

Taking absolute values and the supremum over all finitely simple functions g on Y

with Lq′ norm equal to one, we conclude the proof of (1.3.28) for finitely simple

functions f with Lp norm one. Then (1.3.28) follows by replacing f by f/‖ f‖Lp .

We end this section with the proof of Lemma 1.3.8.

Proof of Lemma 1.3.8. Recall the Poisson integral formula

U(z) =1

2π

∫ +π

−πU(Reiϕ)

R2−ρ2

|Reiϕ −ρeiθ |2 dϕ , z = ρeiθ , (1.3.34)

which is valid for a harmonic function U defined on the unit disk D = z : |z|< 1when |z|< R < 1. See [307, p. 258].

Consider now a subharmonic function u on D that is continuous on the circle

|ζ |= R < 1. When U = u, the right side of (1.3.34) defines a harmonic function on

the set z ∈ C : |z|< R that coincides with u on the circle |ζ |= R. The maximum

principle for subharmonic functions ([307, p. 362]) implies that for |z| < R < 1 we

have

u(z)≤ 1

2π

∫ +π

−πu(Reiϕ)

R2−ρ2

|Reiϕ −ρeiθ |2 dϕ , z = ρeiθ . (1.3.35)

This is valid for all subharmonic functions u on D that are continuous on the circle

|ζ |= R when ρ < R < 1.

It is not difficult to verify that

h(ζ ) =1

πilog

(i1+ζ

1−ζ

)

is a conformal map from D onto the strip S = (0,1)×R. Indeed, i(1+ ζ )/(1− ζ )lies in the upper half-plane and the preceding complex logarithm is a well defined

holomorphic function that takes the upper half-plane onto the strip R×(0,π). Since

F h is a holomorphic function on D, log |F h| is a subharmonic function on D.

Applying (1.3.35) to the function z → log |F(h(z))|, we obtain

log |F(h(z))| ≤ 1

2π

∫ +π

−πlog |F(h(Reiϕ))| R2−ρ2

R2−2ρRcos(θ −ϕ)+ρ2dϕ (1.3.36)

when z = ρeiθ and |z| = ρ < R. Observe that when |ζ | = 1 and ζ = ±1, h(ζ ) has

real part zero or one. It follows from the hypothesis that

log |F(h(ζ ))| ≤ Aeτ0|Imh(ζ )| = Aeτ0

∣∣∣Im 1πi log

(i

1+ζ1−ζ

)∣∣∣= Ae

τ0π

∣∣∣log

∣∣ 1+ζ1−ζ

∣∣ ∣∣∣.

Therefore, log |F(h(ζ ))| is bounded by a multiple of |1+ζ |−τ0/π |1−ζ |−τ0/π , which

is integrable over the set |ζ |= 1, since τ0 < π . Fix now z = ρeiθ with ρ < R and let

R→ 1 in (1.3.36). The Lebesgue dominated convergence theorem gives that

log |F(h(ρeiθ ))| ≤ 1

2π

∫ +π

−πlog |F(h(eiϕ))| 1−ρ2

1−2ρ cos(θ −ϕ)+ρ2dϕ . (1.3.37)

Setting x = h(ρeiθ ), we obtain that

ρeiθ = h−1(x) =eπix− i

eπix + i=−i

cos(πx)

1+ sin(πx)=

(cos(πx)

1+ sin(πx)

)e−i(π/2) ,

from which it follows that ρ = (cos(πx))/(1+ sin(πx)) and θ = −π/2 when 0 <x≤ 1

2, while ρ =−(cos(πx))/(1+ sin(πx)) and θ = π/2 when 1

2≤ x < 1. In either

case we easily deduce that

1−ρ2

1−2ρ cos(θ −ϕ)+ρ2=

sin(πx)

1+ cos(πx)sin(ϕ).

Using this we write (1.3.37) as

log |F(x)| ≤ 1

2π

∫ π

−π

sin(πx)

1+ cos(πx)sin(ϕ)log |F(h(eiϕ))|dϕ . (1.3.38)

We now change variables. On the interval [−π,0) we use the change of variables

it = h(eiϕ) or, equivalently, eiϕ =− tanh(πt)− isech(πt). Observe that as ϕ ranges

from −π to 0, t ranges from +∞ to −∞. Furthermore, dϕ = −π sech(πt)dt. We

have

1

2π

∫ 0

−π

sin(πx)

1+ cos(πx)sin(ϕ)log |F(h(eiϕ))|dϕ

=1

2

∫ ∞

−∞

sin(πx)

cosh(πt)− cos(πx)log |F(it)|dt .

(1.3.39)

On the interval (0,π] we use the change of variables 1+ it = h(eiϕ) or, equivalently,

eiϕ = − tanh(πt)+ isech(πt). Observe that as ϕ ranges from 0 to π , t ranges from

−∞ to +∞. Furthermore, dϕ = π sech(πt)dt. Similarly, we obtain

1

2π

∫ π

0

sin(πt)

1+ cos(πt)sin(ϕ)log |F(h(eiϕ))|dϕ

=1

2

∫ +∞

−∞

sin(πx)

cosh(πt)+ cos(πx)log |F(1+ it)|dt.

(1.3.40)

Adding (1.3.39) and (1.3.40) and using (1.3.38) we conclude the proof when y = 0.

We now consider the case where y = 0. Fix y = 0 and define the function G(z) =F(z+ iy). Then G is analytic on the open strip S = z ∈ C : 0 < Re z < 1 and

continuous on its closure. Moreover, for some A < ∞ and 0≤ τ0 < π we have

log |G(z)|= log |F(z+ iy)| ≤ Aeτ0|Im z+y| ≤ Aeτ0|y| eτ0|Im z|

for all z ∈ S. Then the case y = 0 for G (with A replaced by Aeτ0|y|) yields

|G(x)| ≤ exp

sin(πx)

2

∫ ∞

−∞

[log |G(it)|


log |G(1+ it)|cosh(πt)+cos(πx)

]dt

,

which yields the required conclusion for any real y, since G(x) = F(x+ iy), G(it) =F(it + iy), and G(1+ it) = F(1+ it + iy).

Exercises

1.3.1. Generalize Theorem 1.3.2 to the situation in which T is quasi-subadditive,

that is, it satisfies for some K > 0,

|T ( f +g)| ≤ K(|T ( f )|+ |T (g)|) ,

for all f , g in the domain of T . Prove that in this case, the constant A in (1.3.7) can

be taken to be K times the constant in (1.3.8).

1.3.2. Let (X ,µ), (Y,ν) be two σ -finite measure spaces. Let 1 < p < r ≤ ∞ and

suppose that T be a sublinear operator defined on the space Lp0(X)+Lp1(X) and

taking values in the space of measurable functions on Y . Assume that T maps L1(X)to L1,∞(Y ) with norm A0 and Lr(X) to Lr(Y ) with norm A1. Let 0 < p0 < p1 ≤ ∞.

Prove that T maps Lp to Lp with norm at most

8(p−1)−1p A

1p− 1

r

1− 1r

0 A

1− 1p

1− 1r

1 .

[Hint: First interpolate between L1 and Lr using Theorem 1.3.2 and then interpolate

between Lp+1

2 and Lr using Theorem 1.3.4.]

1.3.3. Let 0 < p0 < p < p1 ≤ ∞ and let T be an operator as in Theorem 1.3.2 that

also satisfies

|T ( f )| ≤ T (| f |) ,for all f ∈ Lp0 +Lp1 .

(a) If p0 = 1 and p1 = ∞, prove that T maps Lp to Lp with norm at most

p

p−1A

1p

0 A1− 1

p

1 .

(b) More generally, if p0 < p <∞, prove that the norm of T from Lp to Lp is at most

p1+ 1

p

[B(p0 +1, p− p0)

pp00 (p− p0)p−p0

] 1p

A

p0p

0 A1− p0

p

1 ,

where B(s, t) =∫ 1

0 xs−1(1− x)t−1 dx is the usual Beta function.

(c) When 0 < p0 < p1 < ∞, then the norm of T from Lp to Lp is at most

min0<λ<1

p1p

(B(p− p0, p0 +1)

(1−λ )p0+

p1−p+1p1−p

λ p1

) 1p

A

1p− 1

p11

p0− 1

p10 A

1p0− 1

p

1p0− 1

p11 .

[Hint: The hypothesis |T ( f )| ≤ T (| f |) reduces matters to nonnegative functions.

Parts (a), (b): Given f ≥ 0 and α > 0 write f = f0 + f1, where f0 = f − λα/A1

when f ≥ λα/A1 and zero otherwise. Here 0 < λ < 1 to be chosen later. Then we

have that ||T ( f )| > α| ≤ ||T ( f0)| > (1− λ )α|. Part (c): Write f = f0 + f1,

where f0 = f −δα when f ≥ δα and zero otherwise. Use that

||T ( f )|> α| ≤ ||T ( f0)|> (1−λ )α|+ ||T ( f1)|> λα|

and optimize over δ > 0.]

1.3.4. Let 0 ≤ γ ,δ < π . For every z ∈ Sa,b = z ∈ C : a < Rez < b, let Tz be a

family of linear operators defined on finetely simple functions on a σ -finite measure

space (X ,µ) and taking values in another σ -finite measure space (Y,ν). Assume that

Tzz is an analytic on of Sa,b, in the sense of (1.3.21), continuous on its closure, and

that for all simple functions f on X and g on Y there is a constant C f ,g <∞ such that

for all z ∈ Sa,b,

log

∣∣∣∣∫

YTz( f )gdν

∣∣∣∣≤C f ,g eγ |Imz|/(b−a) .

Let 1 ≤ p0,q0, p1,q1 ≤ ∞. Suppose that Ta+iy maps Lp0(X) to Lq0(Y ) with bound

M0(y) and Tb+iy maps Lp1(X) to Lq1(Y ) with bound M1(y), where

sup−∞<y<∞

e−δ |y|/(b−a) logM j(y)< ∞ , j = 0,1.

Then for a < t < b, Tt maps Lp(X) to Lq(Y ), where

1

p=

b−tb−a

p0+

t−ab−a

p1and

1

q=

b−tb−a

q0+

t−ab−a

q1.

1.3.5. ([331]) On Rn let x = (x1, . . . ,xn) and |x|= (x21 + · · ·+ x2

n)1/2. Let

Kλ (x) =π

n−12 Γ (λ +1)

Γ (λ + n+12)

∫ +1

−1e2πis|x|(1− s2)λ+

n−12 ds =

Γ (λ +1)

πλ |x|λ+ n2

Jλ+ n2(2π|x|) ,

where λ is a complex number and Jλ+ n2

is the Bessel function of order λ+ n2. Let Tλ

be the operator given by convolution with Kλ . Show that Tλ maps Lp(Rn) to itself

for Reλ > (n−1)| 12− 1

p|.[

Hint: In view of the calculation of the Fourier transform of Kλ contained in Ap-

pendix B.5, we have that when Reλ = 0, Tλ maps L2(Rn) to itself with norm 1. Us-

ing the estimates in Appendices B.6 and B.7, conclude that Kλ is integrable and thus

Tλ maps L1(Rn) to itself with an appropriate constant when Re λ = (n− 1)/2+ δ(for δ > 0). Then use Exercise 1.3.4.

]

1.3.6. Observe that Theorem 1.3.7 yields the stronger conclusion

∥∥Tz( f )∥∥

Lq ≤M(z)∥∥ f

∥∥Lp

for z ∈ S = z ∈ C : 0 < Re z < 1, where for z = x+ iy

M(z) = exp

sin(πx)

2

∫ ∞

−∞

[logM0(t + y)

cosh(πt)− cos(πx)+

logM1(t + y)

cosh(πt)+ cos(πx)

]dt

.

1.3.7. ([380]) Let (X ,µ) and (Y,ν) be two measure spaces with µ(X) < ∞ and

ν(Y ) < ∞. Let T be a countably subadditive operator that maps Lp(X) to Lp(Y )for every 1 0.

(Countably subadditive means that |T (∑ j f j)| ≤ ∑ j |T ( f j)| for all f j in Lp(X) with

∑ j f j ∈ Lp.) Prove that for all f measurable on X we have

∫

Y|T ( f )|dν ≤ 6A(1+ν(Y ))

12

[∫

X| f |(log+2 | f |)α dµ+Cα +µ(X)

12

],

where Cα = ∑∞k=1 kα(2/3)k. This result provides an example of extrapolation.[Hint: Write

f =∞

∑k=0

f χSk,

where Sk = 2k ≤ | f | < 2k+1 when k ≥ 1 and S0 = | f | < 2. Using Holder’s

inequality and the hypotheses on T , obtain that

∫

Y|T ( f χSk

)|dν ≤ 2Aν(Y )1

k+1 2kkαµ(Sk)k

k+1

for k ≥ 1. Note that for k ≥ 1 we have ν(Y )1

k+1 ≤ max(1,ν(Y ))12 and consider the

cases µ(Sk) ≥ 3−k−1 and µ(Sk) ≤ 3−k−1 when summing in k ≥ 1. The term with

k = 0 is easier.]

1.3.8. Prove that for 0 < x < 1 we have

sin(πx)

2

∫ +∞

−∞

1

cosh(πt)+ cos(πx)dt = x ,

sin(πx)

2

∫ +∞

−∞

1

cosh(πt)− cos(πx)dt = 1− x ,

and conclude that Lemma 1.3.8 reduces to Lemma 1.3.5 when the functions M0(y)and M1(y) are constant and assumption (1.3.29) is replaced by the stronger assump-

tion that F is bounded on S.[Hint: In the first integral write cosh(πt) = 1

2(eπt + e−πt). Then use the change of

variables s = eπt .]

1.3.9. Let (X ,µ), (Y,ν) be σ -finite measure spaces, and let 0 < p0 < p1 ≤ ∞. Let

T be a sublinear operator defined on the space Lp0(X)+Lp1(X) and taking values

in the space of measurable functions on Y . Suppose T is a sublinear operator such

that maps Lp0 to L∞ with constant A0 and Lp1 to L∞ with constant A1. Prove T maps

Lp to L∞ with constant 2A1−θ0 Aθ1 where

1−θp0

+θ

p1=

1

p.

1.4 Lorentz Spaces

Suppose that f is a measurable function on a measure space (X ,µ). It would be de-

sirable to have another function f ∗ defined on [0,∞) that is decreasing and equidis-

tributed with f . By this we mean

d f (α) = d f ∗(α) (1.4.1)

for all α ≥ 0. This is achieved via a simple construction discussed in this section.

1.4.1 Decreasing Rearrangements

Definition 1.4.1. Let f be a complex-valued function defined on X . The decreasing

rearrangement of f is the function f ∗ defined on [0,∞) by

f ∗(t) = infs > 0 : d f (s)≤ t= infs≥ 0 : d f (s)≤ t . (1.4.2)

1.4 Lorentz Spaces 49

We adopt the convention inf /0 = ∞, thus having f ∗(t) = ∞ whenever d f (α) > t for

all α ≥ 0. Observe that f ∗ is decreasing and supported in [0,µ(X)].Before we proceed with properties of the function f ∗, we work out three

examples.

f (x)

a3

a2

a1

E3 E1 B2 B3E2 x

a1

a2

a3

0 0 B1 t

f*(t)

.

.

.

.

Fig. 1.3 The graph of a simple function f (x) and its decreasing rearrangement f ∗(t).

Example 1.4.2. Consider the simple function of Example 1.1.2,

f (x) =N

∑j=1

a jχE j(x) ,

where E j are pairwise disjoint sets of finite measure and a1 > · · ·> aN > 0. We saw

in Example 1.1.2 that

d f (α) =N

∑j=0

B jχ[a j+1,a j)(α) ,

where

B j =j

∑i=1

µ(Ei)

and aN+1 = B0 = 0 and a0 = ∞. Observe that for B0 ≤ t < B1, the smallest s > 0

with d f (s)≤ t is a1. Similarly, for B1 ≤ t < B2, the smallest s > 0 with d f (s)≤ t is

a2. Arguing this way, it is not difficult to see that

f ∗(t) =N

∑j=1

a jχ[B j−1,B j)(t) .

See Figure 1.3.

Example 1.4.3. On (Rn,dx) let

f (x) =1

1+ |x|p , 0 < p < ∞ .

A computation shows that

d f (α) =

vn(

1α −1)

np if α < 1 ,

0 if α ≥ 1 ,

and therefore

f ∗(t) =1

(t/vn)p/n +1,

where vn is the volume of the unit ball in Rn.

Example 1.4.4. Again on (Rn,dx) let g(x) = 1− e−|x|2. We can easily see that

dg(α) = 0 if α ≥ 1 and dg(α) = ∞ if α < 1. We conclude that g∗(t) = 1 for all

t ≥ 0. This example indicates that although quantitative information is preserved,

significant qualitative information is lost in passing from a function to its decreasing

rearrangement.

It is clear from the previous examples that f ∗ is continuous from the right and

decreasing. The following are some properties of the function f ∗.

Proposition 1.4.5. For f , g, fn µ-measurable, k ∈C, and 0≤ t,s, t1, t2 <∞ we have

(1) f ∗(d f (α))≤ α whenever α > 0.

(2) d f ( f ∗(t))≤ t.

(3) f ∗(t)> s if and only if t < d f (s); that is, t ≥ 0 : f ∗(t)> s= [0,d f (s)).

(4) |g| ≤ | f | µ-a.e. implies that g∗ ≤ f ∗ and | f |∗ = f ∗.

(5) (k f )∗ = |k| f ∗.(6) ( f +g)∗(t1 + t2)≤ f ∗(t1)+g∗(t2).

(7) ( f g)∗(t1 + t2)≤ f ∗(t1)g∗(t2).

(8) | fn| ↑ | f | µ-a.e. implies f ∗n ↑ f ∗.

(9) | f | ≤ liminfn→∞

| fn| µ-a.e. implies f ∗ ≤ liminfn→∞

f ∗n .

(10) f ∗ is right continuous on [0,∞).

(11) If f ∗(t)<∞, c> 0, and µ(| f | ≥ f ∗(t)−c)<∞, then t ≤ µ(| f | ≥ f ∗(t)).(12) d f = d f ∗ .

(13) (| f |p)∗ = ( f ∗)p when 0 0

tq f ∗(t) = supα>0

α(d f (α)

)qfor 0 < q < ∞ .

Proof. Property (1): The set A = s > 0 : d f (s) ≤ d f (α) contains α and thus

f ∗(d f (α)) = infA≤ α .

Property (2): Let sn ∈ s > 0 : d f (s)≤ t be such that sn ↓ f ∗(t). Then d f (sn)≤ t,

and the right continuity of d f (Exercise 1.1.1 (a)) implies that d f ( f ∗(t))≤ t.

Property (3): If s < f ∗(t) = infu > 0 : d f (u)≤ t, then s /∈ u > 0 : d f (u)≤ twhich gives d f (s)> t. Conversely, if for some t < d f (s) we had f ∗(t)≤ s, applying

d f and using property (2) would yield the contradiction d f (s)≤ d f ( f ∗(t))≤ t.

Properties (4) and (5) are left to the reader.

Properties (6) and (7): Let A= s1 > 0 : d f (s1)≤ t1, B= s2 > 0 : dg(s2)≤ t2,P = s > 0 : d f g(s)≤ t1 + t2, and S = s > 0 : d f+g(s)≤ t1 + t2. Then A+B S

and A ·BP; thus ( f +g)∗(t1+t2) = infS≤ s1+s2 and ( f g)∗(t1+t2) = infP≤ s1s2

are valid for all s1 ∈ A and s2 ∈ B. Taking the infimum over all s1 ∈ A and s2 ∈ B

yields the conclusions.

Property (8): It follows from the definition of decreasing rearrangements that

f ∗n ≤ f ∗n+1 ≤ f ∗ for all n. Let h = limn→∞ f ∗n ; then obviously h ≤ f ∗. Since f ∗n ≤ h,

we have d fn(h(t))≤ d fn( f ∗n (t))≤ t, which implies, in view of Exercise 1.1.1 (c), that

d f (h(t))≤ t by letting n→ ∞. It follows that f ∗ ≤ h, hence h = f ∗.Property (9): Set Fn = infm≥n | fm| and h = liminfn→∞ | fn|= supn≥1 Fn. Since Fn ↑

h, property (8) yields that F∗n ↑ h∗ as n→ ∞. By hypothesis we have | f | ≤ h, hence

f ∗ ≤ h∗ = supn F∗n . Since Fn ≤ | fm| for m≥ n, it follows that F∗n ≤ f ∗m for m≥ n; thus

F∗n ≤ infm≥n f ∗m. Putting these facts together, we obtain f ∗ ≤ h∗ ≤ supn infm≥n f ∗m =liminfn→∞ f ∗n .

Property (10): If f ∗(t0) = 0, then f ∗(t) = 0 for all t > t0 and thus f ∗ is right

continuous at t0. Suppose f ∗(t0)> 0. Pick α such that 0<α < f ∗(t0) and let tn∞n=1

be a sequence of real numbers decreasing to zero. The definition of f ∗ yields that

d f ( f ∗(t0)−α) > t0. Since tn ↓ 0, there is an n0 ∈ Z+ such that d f ( f ∗(t0)−α) >t0 + tn for all n ≥ n0. Property (3) yields that for all n ≥ n0 we have f ∗(t0)−α <f ∗(t0 + tn), and since the latter is at most f ∗(t0), the right continuity of f ∗ follows.

Property (11): The definition of f ∗ yields that the set An = | f | > f ∗(t)− c/nhas measure µ(An) > t. The sets An form a decreasing sequence as n increases and

µ(A1) < ∞ by assumption. Consequently, | f | ≥ f ∗(t) = ⋂∞n=1 An has measure

greater than or equal to t.

Property (12): This is immediate for nonnegative simple functions in view of

Examples 1.1.2 and 1.4.2. For an arbitrary measurable function f , find a sequence of

nonnegative simple functions fn such that fn ↑ | f | and apply (9).

Property (13): It follows from d| f |p(α) = d f (α1/p) = d f ∗(α

1/p) = d( f ∗)p(α) for

all α > 0.

Property (14): This is a consequence of property (12) and of Proposition 1.1.4.

Property (15): This is a restatement of (1.1.2).

Property (16): Given α > 0, without loss of generality we may assume d f (α)> 0.

Pick ε satisfying 0 < ε < d f (α). Property (3) yields f ∗(d f (α)− ε) > α , which

implies that

supt>0

tq f ∗(t)≥ (d f (α)− ε)q f ∗(d f (α)− ε)> (d f (α)− ε)qα .

We first let ε→ 0 and then take the supremum over all α > 0 to obtain one direction.

Conversely, given t > 0, assume without loss of generality that f ∗(t) > 0, and pick

ε such that 0 < ε < f ∗(t). Property (3) yields d f ( f ∗(t)− ε) > t. This implies that

supα>0α(d f (α))q ≥ ( f ∗(t)−ε)(d f ( f ∗(t)−ε))q > ( f ∗(t)−ε)tq. We first let ε→ 0

and then take the supremum over all t > 0 to obtain the opposite direction of the

claimed equality.

1.4.2 Lorentz Spaces

Having disposed of the basic properties of decreasing rearrangements of functions,

we proceed with the definition of the Lorentz spaces.

Definition 1.4.6. Given f a measurable function on a measure space (X ,µ) and

0 < p,q≤ ∞, define

∥∥ f∥∥

Lp,q =

⎧⎪⎪⎨⎪⎪⎩

(∫ ∞

0

(t

1p f ∗(t)

)q dt

t

) 1q

if q < ∞ ,

supt>0

t1p f ∗(t) if q = ∞ .

The set of all f with ‖ f‖Lp,q < ∞ is denoted by Lp,q(X ,µ) and is called the Lorentz

space with indices p and q.

As in Lp and in weak Lp, two functions in Lp,q(X ,µ) are considered equal if they

are equal µ-almost everywhere. Observe that the previous definition implies that

L∞,∞ = L∞, Lp,∞ = weak Lp in view of Proposition 1.4.5 (16) and that Lp,p = Lp.

Remark 1.4.7. Observe that for all 0 < p,r < ∞ and 0 < q≤ ∞ we have

∥∥|g|r∥∥

Lp,q =∥∥g

∥∥r

Lpr,qr . (1.4.3)

On Rn let δ ε( f )(x) = f (εx), ε > 0, be the dilation operator. It is straightforward

that dδ ε ( f )(α) = ε−nd f (α) and (δ ε( f ))∗(t) = f ∗(εnt). It follows that Lorentz norms

satisfy the following dilation identity:

∥∥δ ε( f )∥∥

Lp,q = ε−n/p∥∥ f

∥∥Lp,q . (1.4.4)

Next, we calculate the Lorentz norms of a finitely simple function.

Example 1.4.8. Using the notation of Example 1.4.2, when 0 < p,q < ∞ we have

∥∥ f∥∥

Lp,q =

(p

q

) 1q[

aq1B

qp

1 +aq2

(B

qp

2 −Bqp

1

)+ · · ·+a

qN

(B

qp

N −Bqp

N−1

)] 1q

,

and also ∥∥ f∥∥

Lp,∞ = sup1≤ j≤N

a jB1p

j .

Next, we calculate ‖ f‖L∞,q for the simple function f of Example 1.4.2. when

q < ∞. It turns out that

∥∥ f∥∥

L∞,q=

[a

q1 log

(B1

B0

)+a

q2 log

(B2

B1

)+ · · ·+a

qN log

( BN

BN−1

)] 1q

= ∞ ,

since B0 = 0. We conclude that the only nonnegative simple function with finite L∞,q

norm is the zero function. Given a general nonzero function g∈ L∞,q with 0< q<∞,

there is a nonzero simple function s with 0 ≤ s ≤ g. Then s has infinite norm, and

therefore so does g. We deduce that L∞,q(X) = 0 when 0 < q < ∞.

Proposition 1.4.9. For 0 0 sd f (s)1p when q = ∞ .

(1.4.5)

Proof. The case q = ∞ is statement (16) in Proposition 1.4.5, and we may therefore

concentrate on the case q < ∞. If f is the simple function of Example 1.1.2, then

d f (s) =N

∑j=1

B jχ[a j+1,a j)(s)

with the understanding that aN+1 = 0. Using the this formula and identity in Exam-

ple 1.4.8, we obtain the validity of (1.4.5) for simple functions. In general, given a

measurable function f , find a sequence of nonnegative simple functions such that

fn ↑ | f | a.e. Then d fn ↑ d f (Exercise 1.1.1 (c)) and f ∗n ↑ f ∗ (Proposition 1.4.5 (8)).

Using the Lebesgue monotone convergence theorem we deduce (1.4.5).

Since Lp,p Lp,∞, one may wonder whether these spaces are nested. The next

result shows that for any fixed p, the Lorentz spaces Lp,q increase as the exponent q

increases.

Proposition 1.4.10. Suppose 0 < p ≤ ∞ and 0 < q < r ≤ ∞. Then there exists a

constant cp,q,r (which depends on p, q, and r) such that

∥∥ f∥∥

Lp,r ≤ cp,q,r

∥∥ f∥∥

Lp,q . (1.4.6)

In other words, Lp,q is a subspace of Lp,r.

Proof. We may assume p < ∞, since the case p = ∞ is trivial. We have

t1/p f ∗(t) =

q

p

∫ t

0[s1/p f ∗(t)]q

ds

s

1/q

≤

q

p

∫ t

0[s1/p f ∗(s)]q

ds

s

1/q

since f ∗ is decreasing,

≤(

q

p

)1/q ∥∥ f∥∥

Lp,q .

Hence, taking the supremum over all t > 0, we obtain

∥∥ f∥∥

Lp,∞ ≤(

q

p

)1/q ∥∥ f∥∥

Lp,q . (1.4.7)

This establishes (1.4.6) in the case r = ∞. Finally, when r < ∞, we have

∥∥ f∥∥

Lp,r =

∫ ∞

0[t1/p f ∗(t)]r−q+q dt

t

1/r

≤∥∥ f

∥∥(r−q)/r

Lp,∞

∥∥ f∥∥q/r

Lp,q . (1.4.8)

Inequality (1.4.7) combined with (1.4.8) gives (1.4.6) with cp,q,r = (q/p)(r−q)/rq.

Unfortunately, the functionals ‖ · ‖Lp,q do not satisfy the triangle inequality. For

instance, consider the functions f (t) = t and g(t) = 1− t defined on [0,1]. Then

f ∗(α) = g∗(α) = (1−α)χ[0,1](α). A simple calculation shows that the inequality

‖ f +g‖Lp,q ≤ ‖ f‖Lp,q +‖g‖Lp,q would be equivalent to

p

q≤ 2q Γ (q+1)Γ (q/p)

Γ (q+1+q/p),

which fails in general. However, since for all t > 0 we have

( f +g)∗(t)≤ f ∗(t/2)+g∗(t/2) ,

the estimate ∥∥ f +g∥∥

Lp,q ≤ cp,q

(∥∥ f∥∥

Lp,q +∥∥g

∥∥Lp,q

), (1.4.9)

where cp,q = 21/p max(1,2(1−q)/q), is a consequence of (1.1.4). Also, if ‖ f‖Lp,q = 0

then we must have f = 0 µ-a.e. Therefore, Lp,q is a quasi-normed space for all p,qwith 0 < p,q ≤ ∞. Is this space complete with respect to its quasi-norm? The next

theorem answers this question.

Theorem 1.4.11. Let (X ,µ) be a measure space. Then for all 0 < p,q ≤ ∞, the

spaces Lp,q(X ,µ) are complete with respect to their quasi-norm and they are there-

fore quasi-Banach spaces.

Proof. We consider only the case p < ∞. First we note that convergence in Lp,q

implies convergence in measure. When q = ∞, this is proved in Proposition 1.1.9.

When q < ∞, in view of Proposition 1.4.5 (16) and (1.4.7), it follows that

supt>0

t1/p f ∗(t) = supα>0

αd f (α)1/p ≤

(q

p

)1/q ∥∥ f∥∥

Lp,q

for all f ∈ Lp,q, and from this it follows that convergence in Lp,q implies convergence

in measure.

Now let fn be a Cauchy sequence in Lp,q. Then fn is Cauchy in measure,

and hence it has a subsequence fnk that converges almost everywhere to some f by

Theorem 1.1.13. Fix k0 and apply property (9) in Proposition 1.4.5. Since | f − fnk0|=

limk→∞ | fnk− fnk0

|, it follows that

( f − fnk0)∗(t)≤ liminf

k→∞( fnk

− fnk0)∗(t). (1.4.10)

Raise (1.4.10) to the power q, multiply by tq/p, integrate with respect to dt/t over

(0,∞), and apply Fatou’s lemma to obtain

∥∥ f − fnk0

∥∥q

Lp,q ≤ liminfk→∞

∥∥ fnk− fnk0

∥∥q

Lp,q . (1.4.11)

Now let k0 → ∞ in (1.4.11) and use the fact that fn is Cauchy to conclude that fnk

converges to f in Lp,q. It is a general fact that if a Cauchy sequence has a convergent

subsequence in a quasi-normed space, then the sequence is convergent to the same

limit. It follows that fn converges to f in Lp,q.

Remark 1.4.12. It can be shown that the spaces Lp,q are normable when p, q are

bigger than 1; see Exercise 1.4.3. Therefore, these spaces can be normed to become

Banach spaces.

It is well known that finitely simple functions are dense in Lp of any measure

space, when 0 < p <∞. It is natural to ask whether finitely simple functions are also

dense in Lp,q. This is in fact the case when q = ∞.

Theorem 1.4.13. Finitely simple functions are dense in Lp,q(X ,µ) when 0 < q <∞.

Proof. Let f ∈ Lp,q(X ,µ). Assume without loss of generality that f ≥ 0. Since f

lies in Lp,q Lp,∞ we have µ( f > ε)1/pε ≤ ‖ f‖Lp,q < ∞ for every ε > 0 and

consequently for any A > 0, µ( f > A) is finite and tends to zero as A→ ∞. Thus

for every n = 1,2,3, . . . , there is an An > 0 such that µ( f > An)< 2−n.

For each n = 1,2,3, . . . define the function

ϕn(x) =1+2nAn

∑k=0

k

2nχk2−n< f≤(k+1)2−nχ2−n< f≤An .

Then ϕn is supported in the set 2−n < f ≤ An which has finite µ measure, thus ϕn

is finitely simple and satisfies

f (x)−2−n ≤ ϕn(x)≤ f (x)

for every x ∈ x ∈ X : 2−n < f (x)≤ An. It follows that

µ(x ∈ X : | f (x)−ϕn(x)|> 2−n)< 2−n

which implies that ( f −ϕn)∗(t)≤ 2−n for t ≥ 2−n. Thus

( f −ϕn)∗(t)→ 0 as n→ ∞ and ϕ∗n (t)≤ f ∗(t) for all t > 0.

Since ( f −ϕn)∗(t)≤ f ∗(t), an application of the Lebesgue dominated convergence

theorem gives that ‖ϕn− f‖Lp,q → 0 as n→ ∞.

Remark 1.4.14. One may wonder whether simple functions are dense in Lp,∞. This

turns out to be false for all 0 < p ≤ ∞. However, countable linear combinations of

characteristic functions of sets with finite measure are dense in Lp,∞(X ,µ). We call

such functions countably simple. See Exercise 1.4.4 for details.

1.4.3 Duals of Lorentz Spaces

Given a quasi-Banach space Z with norm ‖ · ‖Z , its dual Z∗ is defined as the space

of all continuous linear functionals T on Z equipped with the norm

∥∥T∥∥

Z∗ = sup‖x‖Z=1

|T (x)| .

Observe that the dual of a quasi-Banach space is always a Banach space.

We are now considering the following question: What are the dual spaces (Lp,q)∗

of Lp,q? The answer to this question presents some technical difficulties for general

measure spaces. In this exposition we restrict our attention to σ -finite nonatomic

measure spaces, where the situation is simpler.

Definition 1.4.15. A measurable subset A of a measure space (X ,µ) is called an

atom if µ(A) > 0 and every measurable subset B of A has measure either equal to

zero or equal to µ(A). A measure space (X ,µ) is called nonatomic if it contains no

atoms. In other words, X is nonatomic if and only if for any A X with µ(A) > 0,

there exists a proper subset B A with µ(B)> 0 and µ(A\B)> 0.

For instance, R with Lebesgue measure is nonatomic, but any measure space

with counting measure is atomic. Nonatomic spaces have the property that every

measurable subset of them with strictly positive measure contains subsets of any

given measure smaller than the measure of the original subset. See Exercise 1.4.5.

Theorem 1.4.16. Suppose that (X ,µ) is a nonatomic σ -finite measure space. Then

(i) (Lp,q)∗ = 0, when 0 < p < 1, 0 < q≤ ∞ ,

(ii) (Lp,q)∗ = L∞, when p = 1, 0 < q≤ 1 ,

(iii) (Lp,q)∗ = 0, when p = 1, 1 < q < ∞ ,

(iv) (Lp,q)∗ = 0, when p = 1, q = ∞ ,

(v) (Lp,q)∗ = Lp′,∞, when 1 < p < ∞, 0 < q≤ 1 ,

(vi) (Lp,q)∗ = Lp′,q′ , when 1 < p < ∞, 1 < q < ∞ ,

(vii) (Lp,q)∗ = 0, when 1 < p < ∞, q = ∞ ,

(viii) (Lp,q)∗ = 0, when p = q = ∞ .

Proof. Since X is σ -finite, we have X =⋃∞

N=1 KN , where KN is an increasing se-

quence of sets with µ(KN) < ∞. Let A be the σ -algebra on which µ is defined

and define AN = A∩KN : A ∈ A . Given T ∈ (Lp,q)∗, where 0 < p,q < ∞, for

each N = 1,2, . . . , consider the measure σN(E) = T (χE) defined on AN . Since σN

satisfies |σN(E)| ≤ (p/q)1/q‖T‖µ(E)1/p, it follows that σN is absolutely continu-

ous with respect to µ restricted on AN . By the Radon–Nikodym theorem (see [153]

(19.36)), there exists a unique (up to a set of µ-measure zero) complex-valued mea-

surable function gN which satisfies∫

KN|gN |dµ < ∞ such that

∫

KN

f dσN =

∫

KN

gN f dµ (1.4.12)

for all f in L1(KN ,AN ,σN). Since σN = σN+1 on AN , it follows that gN = gN+1

µ-a.e. on KN and hence there is a well-defined measurable function g on X that co-

incides with each gN on KN . But the linear functionals f → T ( f ) and f → ∫KN

f dσN

coincide on simple functions supported in KN and therefore they must be equal on

L1(KN ,AN ,σN)∩Lp,q(X ,µ) by density; consequently, (1.4.12) is also equal to T ( f )for f in L1(KN ,AN ,σN)∩Lp,q(X ,µ).

Note that if f ∈ L∞(KN ,µ), then f ∈ Lp,q(KN ,µ) and also in L∞(KN ,σN), which is

contained in L1(KN ,AN ,σN). It follows from (1.4.12) and the preceding discussion

that

T ( f ) =∫

Xg f dµ (1.4.13)

for every f ∈ L∞(KN). We have now proved that for every linear functional T on

Lp,q(X ,µ) with 0 < p,q < ∞ there is a function g satisfying∫

KN|g|dµ < ∞ for all

N = 1,2, . . . such that (1.4.13) holds for all f ∈ L∞(KN).We now examine each case (i)–(viii) separately.

(i) We consider the case 0< p< 1. Let f =∑n anχEn be a finitely simple function

on X (which is taken to be countably simple when q = ∞). Since X is nonatomic,

we split each En as a union of m disjoint sets E j,n, j = 1,2, . . . ,m, each having

measure m−1µ(En). Let f j =∑n anχE j,n . We see that ‖ f j‖Lp,q = m−1/p‖ f‖Lp,q . Now

if T ∈ (Lp,q)∗, it follows that

|T ( f )| ≤m

∑j=1

|T ( f j)| ≤ ‖T‖m

∑j=1

∥∥ f j

∥∥Lp,q = ‖T‖m1−1/p‖ f‖Lp,q .

Let m→ ∞ and use that p < 1 to obtain that T = 0.

(ii) We now consider the case p = 1 and 0 < q≤ 1. Clearly, every h ∈ L∞ gives a

bounded linear functional on L1,q, since

∣∣∣∣∫

Xf hdµ

∣∣∣∣≤∥∥h

∥∥L∞

∥∥ f∥∥

L1 ≤Cq

∥∥h∥∥

L∞

∥∥ f∥∥

L1,q .

Conversely, suppose that T ∈ (L1,q)∗ where q ≤ 1. The function g given in (1.4.13)

satisfies ∣∣∣∣∫

Egdµ

∣∣∣∣=∣∣T (χE)

∣∣≤∥∥T

∥∥q−1/qµ(E)

for all E KN , and hence |g| ≤ q−1/q‖T‖ µ-a.e. on every KN ; see [307, Theorem

1.40 on p. 31] for a proof of this fact. It follows that ‖g‖L∞ ≤ q−1/q‖T‖ and hence

(L1,q)∗ = L∞.

(iii) Let us now take p = 1, 1 < q < ∞, and suppose that T ∈ (L1,q)∗. Then

∣∣∣∣∫

Xf gdµ

∣∣∣∣≤ ‖T‖∥∥ f

∥∥L1,q , (1.4.14)

where g is the function in (1.4.13) and f ∈ L∞(KN). We will show that g = 0 a.e.

Suppose that |g| ≥ δ on some set E0 with µ(E0)> 0. Then there exists N such that

µ(E0 ∩KN) > 0. Let f = g|g|−2χE0∩KNhχh≤M , where h is a nonnegative function.

Then (1.4.14) implies for all h≥ 0 that

∥∥hχh≤M

∥∥L1(E0∩KN)

≤ ‖T‖∥∥hχh≤M

∥∥L1,q(E0∩KN)

.

Letting M → ∞, we obtain that L1,q(E0 ∩ KN) is contained in L1(E0 ∩ KN), but

since the reverse inclusion is always valid, these spaces must be equal. Since X

is nonatomic, this can’t happen; see Exercise 1.4.8 (d). Thus g = 0 µ-a.e. and T = 0.

(iv) In the case p = 1, q = ∞ an interesting phenomenon appears. Since every

continuous linear functional on L1,∞ extends to a continuous linear functional on

L1,q for 1 < q < ∞, it must necessarily vanish on all simple functions by part (iii).

However, (L1,∞)∗ contains nontrivial linear functionals; see [84], [85].

(v) We now take up the case 1 < p < ∞ and 0 < q ≤ 1. Using Exercise 1.4.1 (b)

and Proposition 1.4.10, we see that if f ∈ Lp,q and h ∈ Lp′,∞, then

∫

X| f h|dµ ≤

∫ ∞

0t

1p f ∗(t)t

1p′ h∗(t)

dt

t

≤∥∥ f

∥∥Lp,1‖h‖Lp′,∞

≤Cp,q

∥∥ f∥∥

Lp,q‖h‖Lp′,∞ ;

thus every h ∈ Lp′,∞ gives rise to a bounded linear functional f → ∫h f dµ on Lp,q

with norm at most Cp,q‖h‖Lp′,∞ . Conversely, let T ∈ (Lp,q)∗ where 1 αfor α > 0 and using that

∣∣∣∣∫

Xf gdµ

∣∣∣∣≤ ‖T‖‖ f‖Lp,q ,

we obtain that

αµ(KN ∩|g|> α)≤ (p/q)1/q∥∥T

∥∥µ(KN ∩|g|> α)1p .

Divide by µ(KN ∩|g| > α)1p , let N → ∞, and take the supremum over α > 0 to

obtain that ‖g‖Lp′,∞ ≤ (p/q)1/q‖T‖.

(vi) Using Exercise 1.4.1 (b) and Holder’s inequality, we obtain

∣∣∣∣∫

Xf ϕ dµ

∣∣∣∣≤∫ ∞

0t

1p f ∗(t) t

1p′ ϕ∗(t)

dt

t≤

∥∥ f∥∥

Lp,q

∥∥ϕ∥∥

Lp′,q′ ;

thus every ϕ ∈ Lp′,q′ gives a bounded linear functional on Lp,q with norm at most

‖ϕ‖Lp′,q′ . Conversely, let T be in (Lp,q)∗. By (1.4.13), T is given by integration

against a locally integrable function g. It remains to prove that g ∈ Lp′,q′ . We let

gN,M = gχKNχ|g|≤M . Then (gN,M)∗ ≤ g∗ for all M,N = 1,2, . . . and (gN,M)∗ ↑ g∗ as

M,N → ∞ by Proposition 1.4.5 (4), (8).

For a bounded function f in Lp,q(X) we have

∫ ∞

0f ∗(t)(gN,M)∗(t)dt = sup

h: dh=d f

∣∣∣∣∫

XhgN,M dµ

∣∣∣∣

= suph: dh=d f

∣∣∣∣∫

KN

hχ|g|≤M gdµ

∣∣∣∣

= suph: dh=d f

∣∣T (hχKNχ|g|≤M)

∣∣

≤ suph: dh=d f

∥∥T∥∥ ∥∥hχKN

χ|g|≤M

∥∥Lp,q

≤ suph: dh=d f

∥∥T∥∥ ∥∥h

∥∥Lp,q

=∥∥T

∥∥ ∥∥ f∥∥

Lp,q ,

(1.4.15)

where the first equality is a consequence of the fact that X is nonatomic (see Exercise

1.4.5 (d)). Using the result of Exercise 1.4.5 (b), pick a function f on X such that

f ∗(t) =∫ ∞

t/2s

q′p′−1

(gN,M)∗(s)q′−1 ds

s, (1.4.16)

noting that the preceding integral converges since (gN,M)∗(s) ≤ M χ[0,µ(KN)](s). It

follows that f ∗≤ cp,q Mq′−1, which implies that f is bounded, and also that f ∗(t)= 0

when t > 2µ(KN), which implies that f is supported in a set of measure at most

2µ(KN); thus the function f defined in (1.4.16) is bounded and lies in Lp,q(X).We have the following calculation regarding the Lp,q norm of f :

∥∥ f∥∥

Lp,q =

(∫ ∞

0t

qp

[∫ ∞

t/2s

q′p′−1


s

]qdt

t

) 1q

≤C1(p,q)

(∫ ∞

0(t

1p′ (gN,M)∗(t))q′ dt

t

) 1q

=C1(p,q)∥∥gN,M

∥∥q′/q

Lp′,q′ < ∞ ,

(1.4.17)

which is a consequence of Hardy’s second inequality in Exercise 1.2.8 with b= q/p.

Using (1.4.15) and (1.4.17) we deduce that

∫ ∞

0f ∗(t)(gN,M)∗(t)dt ≤

∥∥T∥∥ ∥∥ f

∥∥Lp,q ≤C1(p,q)

∥∥T∥∥ ∥∥gN,M

∥∥q′−1

Lp′,q′ . (1.4.18)

On the other hand, we have

∫ ∞

0f ∗(t)(gN,M)∗(t)dt ≥

∫ ∞

0

∫ t

t/2s

q′p′−1


s(gN,M)∗(t)dt

≥∫ ∞

0(gN,M)∗(t)q′

∫ t

t/2s

q′p′−1 ds

sdt

= C2(p,q)∥∥gN,M

∥∥q′

Lp′,q′ .

(1.4.19)

Combining (1.4.18) and (1.4.19), and using the fact that ‖gN,M‖Lp′,q′ < ∞, we obtain

‖gN,M‖Lp′,q′ ≤C(p,q)‖T‖. Letting N,M→∞ we deduce ‖g‖Lp′,q′ ≤C(p,q)‖T‖ and

this proves the reverse inequality required to complete case (vi).

(vii) For a complete characterization of this space, we refer to [83].

(viii) The dual of L∞ = L∞,∞ can be identified with the set of all bounded finitely

additive set functions; see [99].

Remark 1.4.17. Some parts of Theorem 1.4.16 are false if X is atomic. For instance,

the dual of ℓp(Z) contains ℓ∞ when 0 < p < 1 and thus it is not equal to 0.

1.4.4 The Off-Diagonal Marcinkiewicz Interpolation Theorem

We now present the main result of this section, the off-diagonal extension of

Marcinkiewicz’s interpolation theorem (Theorem 1.3.2). For a measure space (X ,µ),let S(X) be the space of finitely simple functions on X and S+0 (X) be the subset of

S(X) of functions of the form

n

∑i=m

2−iχAi

where m < n are integers and Ai are subsets of X of finite measure. The sets Ai are

not required to be different nor disjoint; consequently, the sum of two elements in

S+0 (X) also belong to S+0 (X). We define Sreal0 (X) = S+0 (X)−S+0 (X) be the space of

all functions of the form f1− f2, where f1, f2 lie in S+0 (X) and S0(X) be the space

of functions of the form h1 + ih2, where h1,h2 lie in Sreal0 (X).

An operator T defined on S0(X) is called quasi-linear if there is a K ≥ 1 such

that

|T (λ f )|= |λ | |T ( f )| and |T ( f +g)| ≤ K(|T ( f )|+ |T (g)|),

for all λ ∈ C and all functions f , g in S0(X). If K = 1, then T is called sublinear.

Definition 1.4.18. Let T be a linear operator defined on the space of finitely simple

functions S(X) on a measure space (X ,µ) and let 0 < p,q≤ ∞. We say that T is of

restricted weak type (p,q) if

∥∥T (χA)∥∥

Lq,∞ ≤Cµ(A)1/p (1.4.20)

for all measurable subsets A of X with finite measure. Estimates of the form (1.4.20)

are called restricted weak type estimates.

It is important to observe that if an operator is of restricted weak type (p0,q0) and

of restricted weak type (p1,q1), then it is of restricted weak type (p,q), where the

indices are as in (1.4.23). It will be a considerable effort to extend the latter estimate

to all functions in S0(X). The next theorem addresses this extension.

Theorem 1.4.19. Let 0 < r ≤ ∞, 0 < p0 = p1 ≤ ∞, and 0 < q0 = q1 ≤ ∞ and let

(X ,µ), (Y,ν) be σ -finite measure spaces. Let T be a quasi-linear operator defined

on the space of simple functions on X and taking values in the set of measurable

functions on Y . Assume that for some M0,M1 <∞ the following restricted weak type

estimates hold:

∥∥T (χA)∥∥

Lq0 ,∞ ≤M0 µ(A)1/p0 , (1.4.21)

∥∥T (χA)∥∥

Lq1 ,∞≤M1 µ(A)

1/p1 , (1.4.22)

for all measurable subsets A of X with µ(A)< ∞. Fix 0 < θ < 1 and let

1

p=

1−θp0

+θ

p1and

1

q=

1−θq0

+θ

q1. (1.4.23)

Then there exists a constant C∗(p0,q0, p1,q1,K,r,θ)< ∞ such that for all functions

f in S0(X) we have

∥∥T ( f )∥∥

Lq,r ≤C∗(p0,q0, p1,q1,K,r,θ)M1−θ0 Mθ

1

∥∥ f∥∥

Lp,r . (1.4.24)

Additionally, if 0 < p,r < ∞ and if T is linear (or sublinear with nonnegative val-

ues), then it admits a unique bounded extension from Lp,r(X) to Lq,r(Y,ν) such that

(1.4.24) holds for all f in Lp,r.

Before we give the proof of Theorem 1.4.19, we state and prove a lemma that is

interesting on its own.

Lemma 1.4.20. Let 0 < p < ∞ and 0 < q ≤ ∞ and let (X ,µ), (Y,ν) be σ -finite

measure spaces. Let T be a quasi-linear operator defined on S(X) and taking values

in the set of measurable functions on Y . Suppose that there exists a constant M > 0

such that for all measurable subsets A of X of finite measure we have

∥∥T (χA)∥∥

Lq,∞ ≤M µ(A)1p . (1.4.25)

Then for all α with 0 < α < min(q, log2log2K

) there exists a constant C(p,q,K,α)> 0

such that for all functions f in S0(X) we have the estimate

∥∥T ( f )∥∥

Lq,∞ ≤C(p,q,K,α)M∥∥ f

∥∥Lp,α (1.4.26)

where

C(p,q,K,α) = 28+ 2

p+2q K3

(q

q−α

) 2α

(1−2−α)−1α (log2)−

1α .

Proof. A function f in S0(X) can be written as f = h1 − h2 + i(h3− h4), where

h j are in S+0 (X). We write f = f1− f2 + i( f3− f4), where f1 = max(h1− h2,0),f2 = max(−(h1− h2),0), f3 = max(h3− h4,0), and f4 = max(−(h3− h4),0). We

note that f j lie in S+0 (X); indeed, if h1 =∑ℓ 2−ℓχAℓand h2 =∑k 2−kχBk

, where both

sums are finite, then

f1 = ∑ℓ: Aℓ∩(∪kBk)= /0

2−ℓχAℓ+ ∑

(ℓ,k): ℓ<k, Aℓ∩Bk = /0

(2−ℓ−2−k)χAℓ∩Bk.

Since the second sum is equal to ∑ks=ℓ+1 2−sχAℓ∩Bk

, we obtain that h1 ∈ S+0 (X).Likewise we can show that f2, f3, f4 lie in S+0 (X). Moreover, we have f j ≤ | f | and

Proposition 1.4.5(4), yields

‖ f j‖Lp,α (X) ≤ ‖ f‖Lp,α (X)

for all j = 1,2,3,4. Suppose now that (1.4.26) holds for functions in S+0 (X) with

constant C′(p,q,α) in place of C(p,q,K,α). By the quasi-linearity of T we have

‖T ( f )‖Lq,∞(Y ) ≤ K3

∥∥∥∥∥4

∑j=1

|T ( f j)|∥∥∥∥∥

Lq,∞(Y )

≤ K341+ 1

q

4

∑j=1

‖T ( f j)‖Lq,∞(Y )

≤ K342+ 2

q MC′(p,q,K,α)‖ f‖Lp,α (X)

which proves (1.4.26) for all f in S0(X) with constant

C(p,q,K,α) = 24+ 2

q K3 C′(p,q,α) .

We now prove (1.4.26) for functions in S+0 (X) with constant C′(p,q,α) in place

of C(p,q,K,α). It follows from the Aoki–Rolewicz theorem (Exercise 1.4.6) that for

all N ∈ Z+ and for all f1, . . . , fN in S+0 (X) we have the pointwise inequality

|T ( f1 + · · ·+ fN)| ≤ 4

(N

∑j=1

|T ( f j)|α1

) 1α1 ≤ 4

(N

∑j=1

|T ( f j)|α) 1

α

, (1.4.27)

where 0 < α ≤ α1 and α1 satisfies the equation (2K)α1 = 2. The second inequality

in (1.4.27) is a simple consequence of the fact that α ≤ α1. Fix α0 with

0 < α0 ≤ α1 =log2

log2Kand α0 < q .

This ensures that the quasi-normed space Lq/α ,∞ is normable when α ≤ α0. In fact,

since Y is σ -finite, Exercise 1.1.12 gives that the space Ls,∞ is normable as long as

s > 1 and there is an equivalent norm ||| f |||Ls,∞ such that

∥∥ f∥∥

Ls,∞ ≤ f

Ls,∞ ≤

s

s−1

∥∥ f∥∥

Ls,∞ .

Next we claim that for any nonnegative function f in S+0 (X) we have

∥∥T ( f χA)∥∥

Lq,∞ ≤ 4( q

q−α) 1α(1−2−α)−

1α M µ(A)

1p∥∥ f χA

∥∥L∞. (1.4.28)

To show this, we write f =∑nj=m 2− jχS j

, where m < n are integers, S j are subsets of

X of finite measure for all j ∈ m,m+1, . . . ,n, µ(Sm) = 0 and µ(Sn) = 0. Setting

B j = S j ∩A we have

f χA =n

∑j=m

2− jχB j

and 2−m ≤ ‖ f χA‖L∞(X).

We use (1.4.27) once and (1.4.3) twice in the following argument. We have

∥∥T ( f χA)∥∥

Lq,∞ ≤ 4

∥∥∥( n

∑j=m

2− jα |T (χB j)|α

) 1α∥∥∥

Lq,∞

= 4

∥∥∥n

∑j=m


∥∥∥1α

Lq/α,∞

≤ 4

n

∑j=m


1α

Lq/α,∞

≤ 4

(n

∑j=m

2− jα|T (χB j

)|α

Lq/α,∞

) 1α

≤ 4( q

q−α) 1α

(n

∑j=m

2− jα∥∥∥|T (χB j

)|α∥∥∥

Lq/α,∞

) 1α

= 4( q

q−α) 1α

(n

∑j=m

2− jα∥∥T (χB j

)∥∥α

Lq,∞

) 1α

≤ 4( q

q−α) 1α

M

(n

∑j=m

2− jαµ(B j)αp

) 1α

≤ 4( q

q−α) 1α(1−2−α)−

1α M µ(A)

1p 2−m,

using B j A. Using that 2−m ≤ ‖ f χA‖L∞ establishes (1.4.28).

We now apply (1.4.28) to obtain (1.4.26). For any f ∈ S+0 (X) we define measur-

able sets

Ak = x ∈ X : f ∗(2k+1)< | f (x)| ≤ f ∗(2k) (1.4.29)

and we note that these sets are pairwise disjoint. We may write the finitely simple

function f as ∑nj=1 a jχE j

, where 0 < a j <∞, E1 E2 · · · En and 0 < µ(E j)<∞for j ∈ 1,2, . . . ,n. Clearly, we have

f ∗ =n

∑j=1

a jχ[0,µ(E j)).

Thus, when t ∈ (µ(En),∞), f ∗(t) vanishes, and when t ∈ (0,µ(E1)), f ∗(t)=∑nj=1 a j

is a positive constant. So there exists N ∈ Z+ such that f ∗(2k) = 0 when k > N, and

that f ∗(2k) is a positive constant when k < −N. This also implies that Ak = /0 if

|k|> N and thus we express

f =N

∑k=−N

f χAk.


Proposition 1.4.5(2) implies µ(Ak)≤ d f ( f ∗(2k+1))≤ 2k+1. Using (1.4.27) we obtain

∥∥T ( f )∥∥

Lq,∞(Y )≤ 4

∥∥∥( N

∑k=−N

|T ( f χAk)|α

) 1α∥∥∥

Lq,∞(Y )

= 4

∥∥∥N

∑k=−N

|T ( f χAk)|α

∥∥∥1α

Lq/α,∞(Y )

≤ 4

N

∑k=−N

|T ( f χAk)|α

1α

Lq/α,∞(Y )

≤ 4

(N

∑k=−N

|T ( f χAk

)|α

Lq/α,∞(Y )

) 1α

≤ 4( q

q−α) 1α

(N

∑k=−N

∥∥∥|T ( f χAk)|α

∥∥∥Lq/α,∞(Y )

) 1α

≤ 4

(q

q−α

) 1α

(N

∑k=−N

‖T ( f χAk)‖αLq,∞(Y )

) 1α

≤ 16

(q

q−α

) 2α

(1−2−α)−1α M

(N

∑k=−N

µ(Ak)αp∥∥ f χAk

∥∥αL∞

) 1α

≤ 16

(q

q−α

) 2α

(1−2−α)−1α 2

1p M

(∞

∑k=−∞

[ f ∗(2k)]α2kαp

) 1α

≤ 16

(q

q−α

) 2α

(1−2−α)−1α 2

2p (log2)−

1α M‖ f‖Lp,α (X),

where we made use of (1.4.28) and in the last inequality, we used

‖ f‖αLp,α (X) =∞

∑k=−∞

∫ 2k

2k−1tαp [ f ∗(t)]α

dt

t

≥∞

∑k=−∞

(2k−1)αp [ f ∗(2k)]α

∫ 2k

2k−1

dt

t

= 2− α

p log2∞

∑k=−∞

[ f ∗(2k)]α2kαp .

This completes the proof of the required inequality for nonnegative functions in

S+0 (X) with constant

C′(p,q,α) = 16

(q

q−α

) 2α

(1−2−α)−1α 2

2p (log2)−

1α

As noted, the constant in general is C(p,q,K,α) = 24+ 2

p K3 C′(p,q,α).

We now proceed with the proof of Theorem 1.4.19.

Proof. We assume that p0 < p1, since if p0 > p1 we may simply reverse the roles

of p0 and p1. We first consider the case p1,r < ∞. Lemma 1.4.20 implies that

∥∥T ( f )∥∥

Lq0 ,∞ ≤M′0

∥∥ f∥∥

Lp0 ,m ,∥∥T ( f )∥∥

Lq1 ,∞≤M′

1

∥∥ f∥∥

Lp1 ,m,

(1.4.30)

for all f in S0(X), where m = 12

min(q0,q1,

log2log2K

,2r), M′

0 = C(p0,q0,K,m)M0,

M′1 =C(p1,q1,K,m)M1, and C(p,q,K,α) is as in (1.4.26).

Fix a function f in S0(X). Split f = f t + ft as follows:

f t(x) =

f (x) if | f (x)|> f ∗(δ tγ),

0 if | f (x)| ≤ f ∗(δ tγ),

ft(x) =

0 if | f (x)|> f ∗(δ tγ),

f (x) if | f (x)| ≤ f ∗(δ tγ),

where δ is to be determined later and γ is the following nonzero real number:

γ =

1q0− 1

q

1p0− 1

p

=

1q− 1

q1

1p− 1

p1

.

Using Exercise 1.1.10 we write

d f t (v) =

d f (v) when v > f ∗(δ tγ)

d f ( f ∗(δ tγ)) when v≤ f ∗(δ tγ)

d ft (v) =

0 when v≥ f ∗(δ tγ)

d f (v)−d f ( f ∗(δ tγ)) when v < f ∗(δ tγ).

Observe the following facts

v≥ δ tγ =⇒ ( f t)∗(v) ≤ inf

s ∈ (0, f ∗(δ tγ)] : d f t (s)≤ v

= inf

s ∈ (0, f ∗(δ tγ)] : d f ( f ∗(δ tγ))≤ v

= inf(0, f ∗(δ tγ)]

= 0,

v < δ tγ =⇒ ( f t)∗(v) ≤ inf

s > f ∗(δ tγ) : d f t (s)≤ v

= inf

s > f ∗(δ tγ) : d f (s)≤ v

= inf

s > 0 : d f (s)≤ v∩(

f ∗(δ tγ),∞)

= f ∗(v), since f ∗(v)≥ f ∗(δ tγ),

v≥ δ tγ =⇒ ( ft)∗(v) = inf

s > 0 : d ft (s)≤ v

≤ inf

s > 0 : d f (s)≤ v

since d ft ≤ d f

= f ∗(v),

v < δ tγ =⇒ ( ft)∗(v) = inf

s > 0 : d ft (s)≤ v

≤ f ∗(δ tγ), since f ∗(δ tγ) ∈ s > 0 : d ft (s)≤ v

.

We summarize these observations in a couple of inequalities:

( f t)∗(s) ≤

f ∗(s) if 0 < s < δ tγ ,

0 if s≥ δ tγ ,

( ft)∗(s) ≤

f ∗(δ tγ) if 0 < s < δ tγ ,

f ∗(s) if s≥ δ tγ .

It follows from these inequalities that f t lies in Lp0,m and ft lies in Lp1,m for all t > 0.

The quasi-linearity of the operator T and (1.4.9) imply

∥∥T ( f )∥∥

Lq,r

=∥∥t

1q T ( f )∗(t)

∥∥Lr( dt

t )

≤ K∥∥t

1q (|T ( ft)|+ |T ( f t)|)∗(t)

∥∥Lr( dt

t )

≤ K∥∥t

1q T ( ft)

∗( t2)+ t

1q T ( f t)∗( t

2)∥∥

Lr( dtt )

≤ Kar

(∥∥t1q T ( ft)

∗( t2)∥∥

Lr( dtt )+

∥∥t1q T ( f t)∗( t

2)∥∥

Lr( dtt )

)

≤ K max1,2 1r−1

(∥∥t1q T ( ft)

∗( t2)∥∥

Lr( dtt )+

∥∥t1q T ( f t)∗( t

2)∥∥

Lr( dtt )

). (1.4.31)

It follows from (1.4.30) that

t1

q0 T ( f t)∗( t2) ≤ 2

1q0 sup

s>0

s1

q0 T ( f t)∗(s)≤ 21

q0 M′0

∥∥ f t∥∥

Lp0 ,m , (1.4.32)

t1

q1 T ( ft)∗( t

2) ≤ 2

1q1 sup

s>0

s1

q1 T ( ft)∗(s)≤ 2

1q1 M′

1

∥∥ ft∥∥

Lp1 ,m, (1.4.33)

for all t > 0. Now use (1.4.32), (1.4.33), and the facts that

t1q T ( f t)∗( t

2) = t

1q− 1

q0 t1

q0 T ( f t)∗( t2)≤ t

1q− 1

q0 21

q0 M′0

∥∥ f t∥∥

Lp0 ,m

t1q T ( f t)∗( t

2) = t

1q− 1

q1 t1

q1 T ( f t)∗( t2)≤ t

1q− 1

q1 21

q1 M′1

∥∥ f t∥∥

Lp1 ,m,


to estimate (1.4.31) by

K max1,2 1r−1

[2

1q0 M′

0

∥∥∥∥t1q− 1

q0

∥∥ f t∥∥

Lp0 ,m

∥∥∥∥Lr( dt

t )

+21

q1 M′1

∥∥∥∥t1q− 1

q1

∥∥ ft∥∥

Lp1 ,m

∥∥∥∥Lr( dt

t )

],

which is the same as

K max1,2 1r−12

1q0 M′

0

∥∥∥∥t−γ( 1

p0− 1

p )∥∥ f t

∥∥Lp0 ,m

∥∥∥∥Lr( dt

t )

(1.4.34)

+K max1,2 1r−12

1q1 M′

1

∥∥∥∥tγ( 1

p− 1p1

)∥∥ ft∥∥

Lp1 ,m

∥∥∥∥Lr( dt

t )

. (1.4.35)

Next, we change variables u = δ tγ in the Lr quasi-norm in (1.4.34) to obtain

∥∥∥∥t−γ( 1

p0− 1

p )∥∥ f t

∥∥Lp0 ,m

∥∥∥∥Lr( dt

t )

≤ δ1

p0− 1

p

|γ | 1r

∥∥∥∥∥u−( 1

p0− 1

p )(∫ u

0f ∗(s)ms

mp0

ds

s

) 1m

∥∥∥∥∥Lr( du

u )

≤ δ1

p0− 1

p

|γ | 1r

[rm

r( 1p0− 1

p)

] 1m (∫ ∞

0(s

1p0 f ∗(s))rs

−r( 1p0− 1

p ) ds

s

) 1r

=δ

1p0− 1

p

m1m |γ | 1

r ( 1p0− 1

p)

1m

∥∥ f∥∥

Lp,r ,

where the last inequality is a consequence of Hardy’s inequality:

(∫ ∞

0

(∫ u

0g(s)

ds

s

)p

u−b du

u

) 1p

≤ p

b

(∫ ∞

0g(u)p u−b du

u

) 1p

(1.4.36)

with g(s) = f ∗(s)msm/p0 ≥ 0, p = r/m ≥ 1 and b = r/p0− r/p > 0. See Exercise

1.2.8 for the proof of (1.4.36).

Likewise, change variables u = δ tγ in the Lr quasi-norm of (1.4.35) to obtain

∥∥∥∥tγ( 1

p− 1p1

)∥∥ ft∥∥

Lp1 ,m

∥∥∥∥Lr( dt

t )

≤ δ−( 1

p− 1p1

)

|γ | 1r

∥∥∥∥∥u1p− 1

p1

[∫ u

0f ∗(u)ms

mp1

ds

s+

∫ ∞

uf ∗(s)ms

mp1

ds

s

] 1m

∥∥∥∥∥Lr( du

u )

=δ−( 1

p− 1p1

)

|γ | 1r

∥∥∥∥ump− m

p1

∫ u

0f ∗(u)ms

mp1

ds

s+

∫ ∞

uf ∗(s)ms

mp1

ds

s

∥∥∥∥1m

Lr/m( duu )

≤ δ−( 1

p− 1p1

)

|γ | 1r


p1 f ∗(u)m

∫ u

0s

mp1

ds

s

∥∥∥∥Lr/m( du

u )

+


p1

∫ ∞

uf ∗(s)ms

mp1

ds

s

∥∥∥∥Lr/m( du

u )

1m

≤ δ−( 1

p− 1p1

)

|γ | 1r

p1

m

∥∥ f∥∥m

Lp,r +rm

r( 1p− 1

p1)

(∫ ∞

0

(f ∗(u)mu

mp1

) rm u

rp− r

p1du

u

)mr

1m

=δ−( 1

p− 1p1

)

m1m |γ | 1

r

p1p

1p− 1

p1

1m ∥∥ f

∥∥Lp,r ,

where the last inequality above is Hardy’s inequality:

(∫ ∞

0

(∫ ∞

ug(s)

ds

s

)p

ub du

u

) 1p

≤ p

b

(∫ ∞

0g(u)p ub du

u

) 1p

(1.4.37)

with g(s) = f ∗(s)msm/p1 ≥ 0, p = r/m ≥ 1 and b = r/p− r/p1 > 0. See Exercise

1.2.8 for the proof of (1.4.37).

Combining these elements we deduce that given f in S0(X), we have that the

expression in (1.4.34) plus the expression in (1.4.35) is at most

K max1,2 1r−1

m1m |γ | 1

r

⎧⎨⎩

21

q0 M′0 δ

1p0− 1

p

( 1p0− 1

p)

1m

+2

1q1 ( p1

p)

1m M′

1 δ−( 1

p− 1p1

)

( 1p− 1

p1)

1m

⎫⎬⎭

∥∥ f∥∥

Lp,r

We choose δ > 0 such that the two terms in the curly brackets above are equal. We

deduce that

∥∥T ( f )∥∥

Lq,r ≤2K max1,2 1

r−1m

1m |γ | 1

r

⎧⎨⎩

21−θq0 (M′

0)1−θ

( 1p0− 1

p)

1−θm

2θq1 ( p1

p)θm (M′

1)θ

( 1p− 1

p1)θm

⎫⎬⎭

∥∥ f∥∥

Lp,r

where θ is as in (1.4.23), i.e.,

θ =

1p0− 1

p

1p0− 1

p1

.

This proves (1.4.24) in the case p1,r < ∞ with constant C∗(p0,q0, p1,q1,K,r,θ)equal to

2K max1,2 1r−1

m1m |γ | 1

r

⎧⎨⎩

21−θq0 C(p0,q0,K,m)1−θ2

θq1 ( p1

p)θm C(p1,q1,K,m)θ

( 1p0− 1

p)

1−θm ( 1

p− 1

p1)θm

⎫⎬⎭ ,

where we recall that m= 12

min(q0,q1,

log2log2K

,2r)

and C(p j,q j,K,m) is as in Lemma

1.4.20.

We now turn to the remaining cases p = ∞ or r = ∞. The restriction r < ∞ can

be removed since C∗(p0,q0, p1,q1,K,r,θ) has a finite limit as r→∞ and, moreover,

‖ f‖Lp,r = ‖t1/p f ∗(t)‖Lr(dt/t)→‖t1/p f ∗(t)‖L∞(dt/t) = ‖ f‖Lp,∞ as r→∞ and likewise

‖T ( f )‖Lp,r → ‖T ( f )‖Lp,∞ as r → ∞; see Exercise 1.1.3 (a). The restriction p1 < ∞can be removed as follows. Suppose that p1 = ∞. Then, since θ ∈ (0,1) it follows

that p <∞ and we pick p2 > p and p2 <∞. It is easy to see T satisfies the restricted

weak type (p2,q2) estimate

supα>0

αν(|T (χA)|> α)1

q2 ≤M1−ϕ0 M

ϕ1 µ(A)

1p2 ,

where1−ϕ

p0+ϕ

∞=

1

p2,

1−ϕq0

+ϕ

q1=

1

q2. (1.4.38)

Using the result obtained when p1 < ∞ with p2 in place of p1 we obtain that

∥∥T ( f )∥∥

Lq,r ≤C∗(p0,q0, p2,q2,K,r,ρ)M1−ρ0 (M

1−ϕ0 M

ϕ1 )

ρ∥∥ f

∥∥Lp,r (1.4.39)

for all functions f in S0(X), where

1−ρp0

+ρ

p2=

1

p,

1−ρq0

+ρ

q2=

1

q. (1.4.40)

Combining (1.4.38) and (1.4.40) and using (1.4.23) we deduce that θ = ρϕ and

hence (1.4.39) yields (1.4.24) in the case where p1 = ∞. In this case we have

C∗(p0,q0,∞,q1,K,r,θ) =C∗(p0,q0,p0

1−ϕ ,(1−ϕq0

+ ϕq1)−1,K,r, θϕ ) ,

where ϕ is any number satisfying 1 > ϕ > 1− p0p

.

Finally, we address the last assertion of the theorem which claims that when

p,r < ∞ and K = 1, the linear (or sublinear with nonnegative values) operator T

initially defined on finitely simple functions has a unique bounded extension from

Lp,r(X) to Lq,r(Y ), which also satisfies (1.4.24) (with the same constant). To obtain

this conclusion, we will need to know that the space S0(X) is dense in Lp,r(X)whenever 0 < p,r < ∞. This is proved in Proposition 1.4.21 below. Assuming this

proposition, we define the extension of T on Lp,r(X) as follows:

Given f in Lp,r(X) a sequence of functions f j in S0(X) that converge to f in

Lp,r(X), notice that the linearity (or the sublinearity and the fact that T ( f ) ≥ 0 for

all f in S0(X)) implies

|T ( f j)−T ( fk)| ≤ |T ( f j− fk)| .

Using the boundedness of T from Lp,r(X) to Lq,r(Y ) we obtain that the sequence

T ( f j) j is Cauchy in Lq,r(Y ) and by the completeness of this space, it must con-

verge to a limit which we call T ( f ). We observe that T ( f ) is independent of the

choice of the sequence f j j that converges to f in Lp,r. Moreover, one can show

that T is linear (or sublinear with nonnegative values), T ( f ) coincides with T ( f )

on S0(X) and T is bounded from Lp,r(X) to Lq,r(Y ). Thus T is the unique bounded

extension of T on the entire space Lp,r(X). For details, see Exercise 1.4.17.

Proposition 1.4.21. For all 0 < p,r < ∞ the space S0(X) is dense in Lp,r(X).

Proof. Let f ∈ Lp,r(X) and assume first that f ≥ 0. Using (1.4.5) and the fact that

d f is decreasing on [0,∞), we obtain for any n ∈ Z+,

∥∥ f∥∥r

Lp,r(X)= p

∫ ∞

0

[d f (s)

1p s]r ds

s

≥ p

∫ 2−n

0

[d f (2

−n)] r

p sr−1 ds

=p2−nr

r

[d f (2

−n)] r

p ,

which implies that d f (2−n)< ∞. Likewise, again in view of (1.4.5), we have

∥∥ f∥∥r

Lp,r(X)≥ p

∫ 2n

0

[d f (s)

] rp sr−1 ds =

p2nr

r

[d f (2

n)] r

p ,

which implies that limn→∞ d f (2n) = 0. Thus, for any n ∈ Z+, there exists kn ∈ N

such that

d f (2kn) = µ

(x ∈ X : f (x)> 2kn

)< 2−n.

Let En =

x ∈ X : 2−n < f (x)≤ 2kn

and note that µ (En)≤ d f (2−n)< ∞ for each

n ∈ Z+. We write f χEn in binary expansion, that is, f χEn(x) = ∑∞j=−knd j(x)2

− j,

where d j(x) = 0 or 1. Let B j = x ∈ En : d j(x) = 1. Then, µ(B j) ≤ µ(En) and

f χEn can be expressed as f χEn = ∑∞j=−kn2− jχB j

.

Set fn = ∑nj=−kn

2− jχB j. It is obvious that fn ∈ S+0 (X) and fn ≤ f χEn ≤ f . Ob-

serve that when x ∈ En, we have

f (x)− fn(x) =∞

∑j=n+1

2− jχB j≤ 2−n,

and that when x /∈ En, we have fn(x) = 0 and f (x) > 2kn or f (x) ≤ 2−n. It follows

from these facts that

d f− fn(2−n) = µ

(En∩ f − fn > 2−n

)+µ

(Ec

n ∩ f − fn > 2−n)< 2−n.

Hence, for 2−n ≤ t < ∞ one has

( f − fn)∗(t)≤ ( f − fn)

∗(2−n) = inf

s > 0 : d f− fn(s)≤ 2−n≤ 2−n.

This implies that limn→∞( f − fn)∗(t) = 0 for all t ∈ (0,∞). By Proposition 1.4.5 (5),

(6), we obtain for all t ∈ (0,∞)

( f − fn)∗(t)≤ f ∗(t/2)+ f ∗n (t/2)≤ 2 f ∗(t/2).

The Lebesgue dominated convergence theorem gives ‖ fn− f‖Lp,r(X)→ 0 as n→ ∞which yields the required conclusion for nonnegative functions f in Lp,r(X).

For a complex-valued function f ∈ Lp,r(X), we write f = f1− f2 + i( f3− f4),where f j are nonnegative functions in Lp,r(X). By the preceding conclusion, there

exist sequences fj

nn∈Z+ , j = 1,2,3,4, in S+0 (X) such that fj

n → f j in Lp,r(X) as

n→ ∞. Set fn = f 1n − f 2

n + i( f 3n − f 4

n ). Using the fact that ‖ ·‖Lp,r(X) is a quasi-norm

we obtain∥∥ f − fn

∥∥Lp,r(X)

≤C(p,r)4

∑j=1

∥∥ f j− f jn

∥∥Lp,r(X)

which tends to zero as n→ ∞. This completes the proof.

Corollary 1.4.22. Let T be as in the statement of Theorem 1.4.19 and let 0 < p0 =p1 ≤ ∞ and 0 < q0 = q1 ≤ ∞. If T is restricted weak type (p0,q0) and (p1,q1) with

constants M0 and M1, respectively, and for some 0 < θ < 1 we have

1

p=

1−θp0

+θ

p1,

1

q=

1−θq0

+θ

q1,

and p≤ q, then T satisfies the strong type estimate

∥∥T ( f )∥∥

Lq ≤C(p0,q0, p1,q1,θ)M1−θ0 Mθ

1

∥∥ f∥∥

Lp (1.4.41)

for all f in S0(X). Moreover, if T is linear (or sublinear with nonnegative values),

then it has a unique bounded extension from Lp(X ,µ) to Lq(Y,ν) that satisfies es-

timate (1.4.41) for all f ∈ Lp(X) with the constant C(p0,q0, p1,q1,θ) replaced by

C(p0,q0, p1,q1,θ)22/p max(1,21/p−1)2.

Proof. Since θ ∈ (0,1) we must have p,q < ∞. Take r = q in Theorem 1.4.19 and

note that ‖ f‖Lp,r ≤ ‖ f‖Lp since p≤ q = r; see Proposition 1.4.10. The last assertion

follows using Exercise 1.4.17.

We now give examples to indicate why the assumptions p0 = p1 and q0 = q1

cannot be dropped in Theorem 1.4.19.

Example 1.4.23. Let X = Y = R and

T ( f )(x) = |x|−1/2∫ 1

0f (t)dt .

Then α|x : |T (χA)(x)|> α|1/2 = 21/2|A∩ [0,1]| and thus T is of restricted weak

types (1,2) and (3,2). But observe that T does not map L2 = L2,2 to Lq,2. Thus

Theorem 1.4.19 fails if the assumption q0 = q1 is dropped. The dual operator

S( f )(x) = χ[0,1](x)∫ +∞

−∞f (t)|t|−1/2 dt

satisfies α|x : |S(χA)(x)|> α|1/q ≤ c|A|1/2 when q = 1 or 3, and thus it furnishes

an example of an operator of restricted weak types (2,1) and (2,3) that is not L2

bounded. Thus Theorem 1.4.19 fails if the assumption p0 = p1 is dropped.

As an application of Theorem 1.4.19, we give the following strengthening of

Theorem 1.2.13.

We end this chapter with a corollary of the proof of Theorem 1.4.19.

Corollary 1.4.24. Let 1 ≤ r < ∞, 1 ≤ p0 = p1 < ∞, and 0 < q0 = q1 ≤ ∞ and let

(X ,µ) and (Y,ν) be σ -finite measure spaces. Let T be a quasi-linear operator de-

fined on Lp0(X)+Lp1(X) and taking values in the set of measurable functions on Y .

Assume that for some M′0,M

′1 < ∞ the following estimates hold for j = 0,1

∥∥T ( f )∥∥

Lq j ,∞(Y )

≤M′j

∥∥ f∥∥

Lp j (X)

, (1.4.42)

for all functions f ∈ Lp j(X). Fix 0 < θ < 1 and let

1

p=

1−θp0

+θ

p1and

1

q=

1−θq0

+θ

q1. (1.4.43)

Then there exists a constant C∗(p0,q0, p1,q1,K,r,θ)< ∞ such that for all functions

f in Lp(X) we have

∥∥T ( f )∥∥

Lq,p ≤C∗(p0,q0, p1,q1,K,r,θ)(M′0)

1−θ (M′1)θ∥∥ f

∥∥Lp . (1.4.44)

Proof. Since Lp(X) is contained in the sum Lp0(X)+Lp1(X), the operator T is well

defined on Lp(X). Hypothesis (1.4.42) implies that (1.4.30) holds for all f ∈ Lp j ,1.

Repeat the proof of Theorem 1.4.19 starting from (1.4.30) fixing a function f in

Lp(X), m = 1 and r = p. We obtain the required conclusion.

Theorem 1.4.25. (Young’s inequality for weak type spaces) Let G be a locally com-

pact group with left Haar measure λ that satisfies (1.2.12) for all measurable subsets

A of G. Let 1 < p,q,r < ∞ satisfy

1

q+1 =

1

p+

1

r. (1.4.45)

Then there exists a constant Bp,q,r > 0 such that for all f in Lp(G) and g in Lr,∞(G)we have ∥∥ f ∗g

∥∥Lq(G)

≤ Bp,q,r

∥∥g∥∥

Lr,∞(G)

∥∥ f∥∥

Lp(G). (1.4.46)

Proof. We fix 1 < p,q < ∞. Since p and q range in an open interval, we can find

p0 < p < p1, q0 < q < q1, and 0 < θ < 1 such that (1.4.23) and (1.4.45) hold.

Let T ( f ) = f ∗ g, defined for all functions f on G. By Theorem 1.2.13, T extends

to a bounded operator from Lp0 to Lq0,∞ and from Lp1 to Lq1,∞. It follows from

the Corollary 1.4.24 that T extends to a bounded operator from Lp(G) to Lq(G).Notice that since G is locally compact, (G,λ ) is a σ -finite measure space and for

this reason, we were able to apply Corollary 1.4.24.

Exercises

1.4.1. (a) Let g be a nonnegative integrable function on a measure space (X ,µ) and

let A be a measurable subset of X . Prove that

∫

Agdµ ≤

∫ µ(A)

0g∗(t)dt.

(b) (G. H. Hardy and J. E. Littlewood) For f and g measurable on a σ -finite measure

space (X ,µ), prove that

∫

X| f (x)g(x)|dµ(x)≤

∫ ∞

0f ∗(t)g∗(t)dt.

Compare this result to the classical Hardy–Littlewood result asserting that for

a j,b j > 0, the sum ∑ j a jb j is greatest when both a j and b j are rearranged in de-

creasing order (for this see [148, p. 261]).

1.4.2. Let (X ,µ) be a measure space. Prove that if f ∈ Lq0,∞(X)∩Lq1,∞(X) for some

0 < q0 < q1 ≤ ∞, then f ∈ Lq,s(X) for all 0 < s≤ ∞ and q0 < q < q1.

1.4.3. ([164]) Given 0 < p,q < ∞, fix an r = r(p,q)> 0 such that r ≤ 1, r ≤ q and

r < p. Let (X ,µ) be a measure space. For t < µ(X) define

f ∗∗(t) = supµ(E)≥t

(1

µ(E)

∫

E| f |r dµ

)1/r

,

while for t ≥ µ(X) (if µ(X)< ∞) let

f ∗∗(t) =

(1

t

∫

X| f |r dµ

)1/r

.

Also define f

Lp,q =

(∫ ∞

0

(t

1p f ∗∗(t)

)q dt

t

) 1q

.

(The function f ∗∗ and the functional f → ||| f |||Lp,q depend on r.)

(a) Prove that the inequality ((( f +g)∗∗)(t))r ≤ ( f ∗∗(t))r +(g∗∗(t))r is valid for all

t ≥ 0. Since r ≤ q, conclude that the functional f → ||| f |||rLp,q is subadditive and

hence it is a norm when r = 1 (this is possible only if p > 1).

(b) Show that for all f we have

∥∥ f∥∥

Lp,q ≤ f

Lp,q ≤

(p

p− r

)1/r ∥∥ f∥∥

Lp,q .

(c) Conclude that Lp,q(X) is metrizable and normable when 1 < p,q < ∞.

1.4.4. Show that on a measure space (X ,µ) the set of countable linear combinations

of simple functions is dense in Lp,∞(X).(b) Prove that finitely simple functions are not dense in Lp,∞(R) for any 0 < p≤ ∞.[Hint: Part (b): Show that the function h(x) = x−1/pχx>0 cannot be approximated

in Lp,∞ by a sequence of finitely simple functions. Given a finitely simple function s

which is nonzero on a set A with |A|> 0, show that ‖s−h‖Lp,∞ ≥ sup0<λ<|A|1/p λ (λ−p

−|A|)1/p = 1.]

1.4.5. Let (X ,µ) be a nonatomic measure space.

(a) If A0 A1 X , 0 < µ(A1)< ∞, and µ(A0)≤ t ≤ µ(A1), show that there exists

an Et A1 with µ(Et) = t.

(b) Given a nonnegative continuous and decreasing function ϕ on [0,∞) such that

ϕ(t) = 0 whenever t ≥ µ(X), prove that there exists a measurable function f on X

with f ∗(t) = ϕ(t) for all t > 0.

(c) Given A X with 0 < µ(A) < ∞ and g an integrable function on X , show that

there exists a subset A of X with µ(A) = µ(A) such that

∫

A|g|dµ =

∫ µ(A)

0g∗(s)ds.

(d) If X is σ -finite, f ∈ L∞(X), and g ∈ L1(X), prove that

suph: dh=d f

∣∣∣∣∫

Xhgdµ

∣∣∣∣=∫ ∞

0f ∗(s)g∗(s)ds,

where the supremum is taken over all functions h on X equidistributed with f .[Hint: Part (a): Reduce matters to the situation in which A0 = /0. Consider first the

case that for all A X there exists a subset B of X satisfying 110µ(A) ≤ µ(B) ≤

910µ(A). Then we can find subsets of A1 of measure in any arbitrarily small inter-

val, and by continuity the required conclusion follows. Next consider the case in

which there is a subset A1 of X such that every B A1 satisfies µ(B) < 110µ(A1)

or µ(B)> 910µ(A1). Without loss of generality, normalize µ so that µ(A1) = 1. Let

µ1 = supµ(C) : C A1, µ(C)< 110 and pick B1 A1 such that 1

2µ1 ≤ µ(B1)≤

µ1. Set A2 = A1 \B1 and define µ2 = supµ(C) : C A2, µ(C)< 110. Continue in

this way and define sets A1 A2 A3 · · · and numbers 110≥ µ1 ≥ µ2 ≥ µ3 ≥ ·· · .

If C An+1 with µ(C)< 110

, then C∪Bn An with µ(C∪Bn)<15< 9

10, and hence

by assumption we must have µ(C∪Bn) <1

10. Conclude that µn+1 ≤ 1

2µn and that

µ(An) ≥ 45

for all n = 1,2, . . . . Then the set⋂∞

n=1 An must be an atom. Part (b):

First show that when d is a simple right continuous decreasing function on [0,∞)there exists a measurable f on X such that f ∗ = d. For general continuous func-

tions, use approximation. Part (c): Let t = µ(A) and define A1 = x : |g(x)|> g∗(t)and A2 = x : |g(x)| ≥ g∗(t). Then A1 A2 and µ(A1) ≤ t ≤ µ(A2). Pick A such

that A1 A A2 and µ(A) = t = µ(A) by part (a). Then∫

Agdµ =

∫X gχ

Adµ =

∫ ∞0 (gχ

A)∗ ds =

∫ µ(A)0 g∗(s)ds. Part (d): Reduce matters to functions f ,g ≥ 0. Let

f = ∑Nj=1 a jχA j

where a1 > a2 > · · · > aN > 0 and the A j are pairwise disjoint.

Write f as ∑Nj=1 b jχB j

, where b j = (a j−a j+1) and B j = A1∪·· ·∪A j. Pick B j as in

part (c). Then B1 · · · BN and the function f1 = ∑Nj=1 b jχB j

has the same distri-

bution function as f . It follows from part (c) that∫

X f1gdµ =∫ ∞

0 f ∗(s)g∗(s)ds. The

case of a general f ∈ L∞(X) follows by approximation by finitely simple functions.]

1.4.6. ([7], [297]) Let K ≥ 1 and let ‖ · ‖ be a nonnegative functional on a vector

space X that satisfies

‖x+ y‖ ≤ K(‖x‖+‖y‖

)

for all x,y ∈ X . For a fixed α ≤ 1 satisfying (2K)α = 2 show that

‖x1 + · · ·+ xn‖α ≤ 4(‖x1‖α + · · ·+‖xn‖α)

for all n = 1,2, . . . and all x1, x2, . . . , xn in X . This inequality is referred to as the

Aoki-Rolewicz theorem.[Hint: Quasi-linearity implies that ‖x1 + · · ·+ xn‖ ≤ max1≤ j≤n[(2K) j‖x j‖] for all

x1, . . . ,xn in X (use that K ≥ 1). Define H : X → R by setting H(0) = 0 and

H(x)= 2 j/α if 2 j−1 < ‖x‖α ≤ 2 j. Then ‖x‖≤H(x)≤ 21/α‖x‖ for all x∈X . Prove by

induction that ‖x1+ · · ·+xn‖α ≤ 2(H(x1)α+ · · ·+H(xn)

α). Suppose that this state-

ment is true when n = m. To show its validity for n = m+1, without loss of general-

ity assume that ‖x1‖≥ ‖x2‖≥ · · · ≥ ‖xm+1‖. Then H(x1)≥H(x2)≥ ·· · ≥H(xm+1).Assume that all the H(x j)’s are distinct. Then since H(x j)

α are distinct powers of 2,

they must satisfy H(x j)α ≤ 2− j+1H(x1)

α . Then

‖x1 + · · ·+ xm+1‖α ≤[

max1≤ j≤m+1

(2K) j‖x j‖]α

≤[

max1≤ j≤m+1

(2K) jH(x j)]α

≤[

max1≤ j≤m+1

(2K) j21/α2− j/αH(x1)]α

= 2H(x1)α

≤ 2(H(x1)α + · · ·+H(xm+1)

α) .

We now consider the case that H(x j) = H(x j+1) for some 1≤ j≤m. Then for some

integer r we must have 2r−1 < ‖x j+1‖α ≤ ‖x j‖α ≤ 2r and H(x j) = 2r/α . Next note

that

‖x j + x j+1‖α ≤ Kα(‖x j‖+‖x j+1‖)α ≤ Kα(2 2r/α)α = 2r+1.

This implies

H(x j + x j+1)α ≤ 2r+1 = 2r +2r = H(x j)

α +H(x j+1)α .

Now apply the inductive hypothesis to x1, . . . ,x j−1,x j + x j+1,x j+1, . . . ,xm and use

the previous inequality to obtain the required conclusion.]

1.4.7. (a) ([347]) Let (X ,µ) and (Y,ν) be measure spaces. Let Z be a Banach space

of complex-valued measurable functions on Y . Assume that Z is closed under abso-

lute values and satisfies ‖ f‖Z = ‖| f |‖Z . Suppose that T is a linear operator defined

on the space of finitely simple functions on (X ,µ) and taking values in Z. Suppose

that for some constant A > 0 the following restricted weak type estimate

∥∥T (χE)∥∥

Z≤ Aµ(E)1/p

holds for some 0 < p < ∞ and for all E measurable subsets of X of finite measure.

Show that for all finitely simply functions f on X we have

∥∥T ( f )∥∥

Z≤ p−1A

∥∥ f∥∥

Lp,1 .

Consequently T has a bounded extension from Lp,1(X) to Z.

(b) ([172]) As an application of part (a) prove that for any U , V measurable subsets

of Rn with |U |, |V |< ∞ and any f measurable on U×V we have

(∫

U

∥∥ f (u, ·)∥∥2

L2,1(V )du

) 12

≤ 1

2

∥∥ f∥∥

L2,1(U×V ).

[Hint: Part (a): Let f = ∑N

j=1 a jχE j≥ 0, where a1 > a2 > · · ·> aN > 0, µ(E j)< ∞

pairwise disjoint. Let Fj = E1 ∪ ·· · ∪E j, B0 = 0, and B j = µ(Fj) for j ≥ 1. Write

f = ∑Nj=1(a j−a j+1)χFj

, where aN+1 = 0. Then

∥∥T ( f )∥∥

Z=

∥∥ |T ( f )|∥∥

Z

≤N

∑j=1

(a j−a j+1)∥∥T (χFj

)∥∥

Z

≤ AN

∑j=1

(a j−a j+1)(µ(Fj))1/p

= AN−1

∑j=0

a j+1(B1/p

j+1−B1/p

j )

= p−1A∥∥ f

∥∥Lp,1 ,

where the penultimate equality follows by a summation by parts; see Appendix F.]

1.4.8. Let 0 < p,q,α,β < ∞. Also let 0 < q1 < q2 < ∞.

(a) Show that the function fα ,β (t) = t−α(log t−1)−β χ[0,e−β/α )(t) lies in Lp,q(R) if

and only if either p < 1/α or both p = 1/α and q > 1/β hold. Conclude that the

function t → t−1/p(log t−1)−1/q1χ[0,e−p/q1 )

(t) lies in Lp,q2(R) but not in Lp,q1(R).

(b) Find a necessary and sufficient condition in terms of p,α,β for the function

gα ,β (t) = (1+ t)−α(log(2+ t))−βχ[0,∞) to lie in Lp,q(R).

(c) Let ψ(t) be smooth decreasing function on [0,∞) and let F(x) = ψ(|x|) for x

in Rn, where |x| is the modulus of x. Show that F∗(t) = f ((t/vn)1/n), where vn is

the volume of the unit ball. Use this formula to construct examples showing that

Lp,q1(Rn) Lp,q2(Rn).(d) On a general nonatomic measure space (X ,µ) prove that there does not exist a

constant C(p,q1,q2)> 0 such that for all f in Lp,q2(X) the following is valid:

∥∥ f∥∥

Lp,q1≤C(p,q1,q2)

∥∥ f∥∥

Lp,q2.

[Hint: Parts (a), (b): Use that fα ,β and gα ,β are equal to their decreasing rearrange-

ments. Part (d): Use Exercise 1.4.5 (b) with ϕ(t) = g1/p,1/q1(t).

]

1.4.9. ([346]) Let Lp(ω) denote the space of all measurable functions f on Rn such

that ‖ f‖p

Lp(ω)=

∫Rn | f (x)|pω(x)dx <∞, where 0 < ω <∞ a.e. Let T be a sublinear

operator that maps Lp0(ω0) to Lq0,∞(ω) and Lp1(ω1) to Lq1,∞(ω), where ω0,ω1,ωare positive functions and 1≤ p0 < p1 < ∞, 0 < q0,q1 < ∞. Suppose that

1

pθ=

1−θp0

+θ

p1,

1

qθ=

1−θq0

+θ

q1.

Let Ωθ = ω1−θp0

pθ

0 ωθp1

pθ

1 . Show that T maps Lpθ(Ωθ

)→ Lqθ ,pθ (ω) .

[Hint: Define L( f ) = (ω1/ω0)

1p1−p0 f and observe that for each θ ∈ [0,1], L maps

Lpθ(Ωθ

)→ Lpθ

((ω p1

0 ω−p01 )

1p1−p0

)isometrically. Then apply Corollary 1.4.24 to

the sublinear operator T L−1.]

1.4.10. ([185], [349]) Let λn be a sequence of positive numbers with ∑nλn ≤ 1 and

∑nλn log( 1λn) = K < ∞. Suppose all sequences are indexed by a fixed countable set.

(a) Let fn be a sequence of complex-valued functions in L1,∞(X) with ‖ fn‖L1,∞ ≤ 1

uniformly in n. Prove that ∑nλn fn lies in L1,∞(X) with norm at most 2(K+2). (This

property is referred to as the logconvexity of L1,∞.)

(b) Let Tn be a sequence of sublinear operators that map L1(X) to L1,∞(Y ) with

norms ‖Tn‖L1→L1,∞ ≤ B uniformly in n. Use part (a) to prove that ∑nλnTn maps

L1(X) to L1,∞(Y ) with norm at most 2B(K +2).(c) Given δ > 0 pick 0 < ε < δ and use the simple estimate

µ(∞

∑n=1

2−δn fn>α)≤

∞

∑n=1

µ(2−δn fn>(2ε −1)2−εnα

)

to obtain a simple proof of the statement in part (a) when λn = 2−δn, n = 1,2, . . . .[Hint: Part (a): For fixed α > 0, write fn = un + vn +wn, where un = fnχ| fn|≤ α

2,

vn = fnχ| fn|> α2λn

, and wn = fnχ α2 <| fn|≤ α

2λn. Let u = ∑nλnun, v = ∑nλnvn, and w =

∑nλnwn. Clearly |u| ≤ α2

. Also v = 0 ⋃n| fn|> α

2λn; hence µ(v = 0)≤ 2

α .

Finally,

∫

X|w|dµ ≤ ∑

n

λn

∫

X| fn|χ α

2 <| fn|≤ α2λn

dµ

≤ ∑n

λn

[∫ α/(2λn)

α/2d fn(β )dβ +

∫ α/2

0d fn(α/2)dβ

]

≤ K +1 .

Using µ(|u+ v+w|> α)≤ µ(|u|> α/2)+µ(|v| = 0)+µ(|w|> α/2),deduce the conclusion.

]

1.4.11. Let fnn be a sequence of measurable functions on a measure space (X ,µ).Let 0 < q,s≤ ∞.

(a) Suppose that fn ≥ 0 for all n. Show that

∥∥ liminfn→∞

fn

∥∥Lq,s ≤ liminf

n→∞

∥∥ fn

∥∥Lq,s .

(b) Let gn → g in Lq,s as n→ ∞. Show that ‖gn‖Lq,s →‖g‖Lq,s as n→ ∞.

1.4.12. (a) Suppose that X is a quasi-Banach space and let X∗ be its dual (which is

always a Banach space). Prove that for all T ∈ X∗ we have

∥∥T∥∥

X∗ = supx∈X‖x‖X≤1

|T (x)| .

(b) Now suppose that X is a Banach space. Use the Hahn–Banach theorem to prove

that for every x ∈ X we have

‖x‖X = supT∈X∗‖T‖X∗≤1

|T (x)| .

Observe that this result may fail for quasi-Banach spaces. For example, if X = L1,∞,

every linear functional on X∗ vanishes on the set of simple functions.

(c) Let 1 < p < ∞, X = Lp,1(Y ), and X∗ = Lp′,∞(Y ), where (Y,µ) is nonatomic

σ -finite measure space. Conclude that

∥∥ f∥∥

Lp,1 ≈ sup‖g‖

Lp′,∞≤1

∣∣∣∣∫

Yf gdµ

∣∣∣∣ ,

∥∥ f∥∥

Lp,∞ ≈ sup‖g‖

Lp′,1≤1

∣∣∣∣∫

Yf gdµ

∣∣∣∣ .

1.4.13. Let 0 < p,q < ∞. Prove that any function in Lp,q(X ,µ) can be written as

f =+∞

∑n=−∞

cn fn,

where fn is a function bounded by 2−n/p, supported on a set of measure 2n, and the

sequence ckk lies in ℓq and satisfies

2− 1

p (log2)1q∥∥ckk

∥∥ℓq ≤

∥∥ f∥∥

Lp,q ≤∥∥ckk

∥∥ℓq 2

1p (log2)

1q .

[Hint: Let cn = 2n/p f ∗(2n), An = x : f ∗(2n+1) < | f (x)| ≤ f ∗(2n)), and fn =

c−1n f χAn .

]

1.4.14. (T. Tao) Let 0 0, and let f be a measurable func-

tion on a measure space (X ,µ).(a) Suppose that ‖ f‖Lp,∞ ≤ A. Then for every measurable set E of finite measure

there exists a measurable subset E ′ of E with µ(E ′) ≥ γ µ(E) such that f is inte-

grable on E ′ and ∣∣∣∣∫

E ′f dµ

∣∣∣∣≤ (1−γ)−1/p Aµ(E)1− 1p .

(b) Suppose that (X ,µ) is a σ -finite measure space and that f has the property that

for any measurable subset E of X with µ(E)< ∞ there is a measurable subset E ′ of

E with µ(E ′)≥ γ µ(E) such that f is integrable on E ′ and

∣∣∣∣∫

E ′f dµ

∣∣∣∣≤ Bµ(E)1− 1p .

Then we have that ‖ f‖Lp,∞ ≤ B41/pγ−1√

2.

(c) Conclude that if (X ,µ) is a σ -finite measure space then

∥∥ f∥∥

Lp,∞ ≈ supEX

0<µ(E)<∞

infE ′E

µ(E ′)≥ 12 µ(E)

f∈L1(E ′)

µ(E)−1+ 1p

∣∣∣∣∫

E ′f dµ

∣∣∣∣ .

[Hint: Part (a): Take E ′ = E \ | f | > A(1− γ)−

1p µ(E)−

1p . Part (b): Write X =⋃∞

n=1 Xn with µ(Xn)< ∞. Given α > 0, note that the set| f |> α

is contained in

Re f > α√

2

∪

Im f > α√2

∪

Re f <− α√2

∪

Im f <− α√2

.

Let En be any of the preceding four sets intersected with Xn, let E ′n be a subset of

it with measure at least γ µ(En) as in the hypothesis. Then∣∣∫

E ′nf dµ

∣∣≥ α√2γ µ(En),

from which it follows that αµ(En)1/p ≤ B

√2γ−1, and let n→ ∞.

]

1.4.15. Let T be a linear operator defined on the set of finitely simple functions on

a σ -finite measure space (X ,µ) and taking values in the set of measurable functions

on a σ -finite measure space (Y,ν) and T t be a linear operator defined on the set

of finitely simple functions on (Y,ν) and taking values in the set of measurable

functions of (X ,µ). Suppose that for all A subsets of X and B subsets of Y of finite

measure we have

∫

B|T (χA)|dν+

∫

A|T t(χB)|dµ < ∞

and that T and T t are related via the “transpose identity”

∫

YT (χA)χB dν =

∫

XT t(χB)χA dµ =Λ(A,B) .

Assume that whenever µ(An)+ν(Bn)→ 0 as n→∞, we have Λ(An,Bn)→ 0. Sup-

pose that T and T t are restricted weak type (1,1) operators, with constants C1 and

C2, respectively. Show that, for all 1 < p < ∞, T is of restricted weak type (p, p).Precisely, show that there exists a constant Kp such that

∥∥T (χA)∥∥

Lp(Y )≤ Kp C

1p

1 C1− 1

p

2 µ(A)1p

for all measurable subsets A of X with µ(A)< ∞.[Hint: Suppose that C1µ(F) > C2ν(E) and pick m so that C1µ(F) ∼ 2mC2ν(E).

Since T t is restricted weak type (1,1) there is an F ′ F such that µ(F ′)≥ 12µ(F)

and |Λ(F ′,E)| ≤ 2C2ν(E). Find by induction sets F( j) F \ (F ′ ∪ ·· · ∪ F( j−1))such that µ(F( j)) ≥ 1

2µ(F \ (F ′ ∪ ·· · ∪F( j−1))) and |Λ(F( j),E)| ≤ 2C2ν(E), j =

1,2, . . . ,m . Stop when F(m) = F \(F ′∪·· ·∪F(m−1)) satisfies C1µ(F(m))≤C2ν(E).

Since T is restricted weak type (1,1) there is a subset E ′ of E such that ν(E ′) ≥12ν(E) and |Λ(F(m),E ′)| ≤ 2C1µ(F

(m))≤ 2C2ν(E). Now write

Λ(F,E) =m−1

∑j=1

Λ(F( j),E)+Λ(F(m),E ′)+Λ(F(m),E \E ′)

from which it follows that

|Λ(F,E)| ≤ 2C2ν(E)

(1+ log2

C1µ(F)

C2ν(E)

)+ |Λ(F1,E1)|

where F1 = F(m) and E1 = E \ E ′. Note that the first term in the sum above is

at most K′p(C1µ(F))1/p(C2ν(E))1/p′ and that the identical estimate holds if the

roles of E and F are reversed. Also observe that µ(F1) ≤ 12µ(F) and ν(E1) ≤

12ν(E). Continuing this process we find sets (Fn,En) with µ(Fn+1) ≤ 1

2µ(Fn) and

ν(En+1) ≤ 12ν(En).Using Λ(Fn,En) → 0 as n → ∞ we deduce that |Λ(F,E)| ≤

2K′p(C1µ(F))1/p(C2ν(E))1/p′ . Considering the sets E+ = E ∩ T (χF) > 0 and

E− = E ∩ T (χF) < 0, obtain that∫

E

∣∣T (χF)∣∣dν ≤ 4K′p

(C1µ(F)

) 1p(C2ν(E)

) 1p′

for all F and E measurable sets of finite measure. Exercise 1.1.12 (a) with r = 1

yields that ‖T (χF)‖Lp,∞ ≤ 4Kp C1/p

1 C1/p′2 µ(F)1/p.

]

1.4.16. ([35]) Let 0 < p0 < p1 <∞ and 0 < α,β ,A,B <∞. Suppose that a family of

sublinear operators Tk is of restricted weak type (p0, p0) with constant A2−kα and

of restricted weak type (p1, p1) with constant B2kβ for all k ∈ Z. Show that there

is a constant C =C(α,β , p0, p1) such that ∑k∈Z Tk is of restricted weak type (p, p)with constant C A1−θBθ , where θ = α/(α+β ) and

1

p=

1−θp0

+θ

p1.

[Hint: Estimate µ(|T (χE)|> λ) by the sum ∑k≥k0

µ(|Tk(χE)|> cλ2α′(k0−k))+

∑k≤k0µ(|Tk(χE)| > cλ2β

′(k−k0)), where c is a suitable constant and 0 < α ′ < α ,

0 < β ′ < β . Apply the restricted weak type (p0, p0) hypothesis on each term of the

first sum, the restricted weak type (p1, p1) hypothesis on each term of the second

sum, and choose k0 to optimize the resulting expression.]

1.4.17. Let (X ,µ), (Y,ν) be measure spaces, 0 < p,r,q,s≤ ∞ and 0 < B < ∞. Sup-

pose that a sublinear operator T is defined on a dense subspace D of Lp,r(X), takes

values in the space of measurable functions of another measure space Y , and satisfies

T ( f )≥ 0 for all f in D . Assume that

∥∥T (ϕ)∥∥

Lq,s ≤ B∥∥ϕ

∥∥Lp,r

for all ϕ in D . Prove that T admits a unique sublinear extension T on Lp,r(X) such

that ∥∥T ( f )∥∥

Lq,s ≤ B∥∥ f

∥∥Lp,r

for all f ∈ Lp,r(X).[Hint: Given f ∈ Lp,r(X) find a sequence of functions ϕ j in D such that ϕ j → f in

Lp,r. Use the inequality |T (ϕ j)−T (ϕk)| ≤ |T (ϕ j−ϕk)| , to obtain that the sequence

T (ϕ j) j is Cauchy in Lq,s and thus it has a unique limit T ( f ) which is independent

of the choice of sequence ϕ j. Boundedness of T follows by density. To prove that T

is sublinear use that convergence in Lq,s implies convergence in measure and thus a

subsequence of T (ϕ j) converges ν-a.e. to T ( f ). Also use Exercise 1.4.11.]

HISTORICAL NOTES

The modern theory of measure and integration was founded with the publication of Lebesgue’s

dissertation [214]; see also [215]. The theory of the Lebesgue integral reshaped the course of in-

tegration. The spaces Lp([a,b]), 1 < p < ∞, were first investigated by Riesz [290], who obtained

many important properties of them. A rigorous treatise of harmonic analysis on general groups

can be found in the book of Hewitt and Ross [152]. The best possible constant Cpqr in Young’s

inequality ‖ f ∗ g‖Lr(Rn) ≤ Cpqr‖ f‖Lp(Rn)‖g‖Lq(Rn),1p+ 1

q= 1

r+ 1, 1< p,q,r<∞, was shown by

Beckner [21] to be Cpqr = (BpBqBr′ )n, where B2

p = p1/p(p′)−1/p′ .

Theorem 1.3.2 first appeared without proof in Marcinkiewicz’s brief note [240]. After his death

in World War II, this theorem seemed to have escaped attention until Zygmund reintroduced it

in [387]. This reference presents the more difficult off-diagonal version of the theorem, derived

by Zygmund. Stein and Weiss [347] strengthened Zygmund’s theorem by assuming that the initial

estimates are of restricted weak type whenever 1 ≤ p0, p1,q0,q1 ≤ ∞. The extension of this result

to the case 0 < p0, p1,q0,q1 < 1 in Theorem 1.4.19 is due to the author. The critical Lemma 1.4.20

was suggested by Kalton. Improvements of these results, in particular, the appearance of the space

S0(X) and the presence of the factor M1−θ0 Mθ

1 in (1.4.24) appeared in Liang, Liu, and Yang [224].

Equivalence of restricted weak type (1,1) and weak type (1,1) properties for certain maximal

multipliers was obtained by Moon [257]. The following partial converse of Theorem 1.2.13 is due

to Stepanov [351]: If a convolution operator maps L1(Rn) to Lq,∞(Rn) for some 1 < q < ∞ then its

kernel must be in Lq,∞.

The extrapolation result of Exercise 1.3.7 is due to Yano [380]; see also Zygmund [389, pp.

119–120] and the related work of Carro [56], Soria [330], and Tao [356].

The original version of Theorem 1.3.4 was proved by Riesz [293] in the context of bilinear

forms. This version is called the Riesz convexity theorem, since it says that the logarithm of the

function M(α ,β ) = infx,y

∣∣∑nj=1∑

mk=1 a jkx jyk

∣∣‖x‖−1

ℓ1/α ‖y‖−1

ℓ1/β (where the infimum is taken over all

sequences x jnj=1 in ℓ1/α and ykm

k=1 in ℓ1/β ) is a convex function of (α ,β ) in the triangle 0 ≤α ,β ≤ 1, α + β ≥ 1. Riesz’s student Thorin [360] extended this triangle to the unit square 0 ≤α ,β ≤ 1 and generalized this theorem by replacing the maximum of a bilinear form with the

maximum of the modulus of an entire function in many variables. After the end of World War II,

Thorin published his thesis [361], building the subject and giving a variety of applications. The

original proof of Thorin was rather long, but a few years later, Tamarkin and Zygmund [354] gave

a very elegant short proof using the maximum modulus principle in a more efficient way. Today,

this theorem is referred to as the Riesz–Thorin interpolation theorem.

Calderon [42] elaborated the complex-variables proof of the Riesz–Thorin theorem into a gen-

eral method of interpolation between Banach spaces. The complex interpolation method can also be

defined for pairs of quasi-Banach spaces, although certain complications arise in this setting; how-

ever, the Riesz–Thorin theorem is true for pairs of Lp spaces (with the “correct” geometric mean

constant) for all 0 < p≤ ∞ and also for Lorentz spaces. In this setting, duality cannot be used, but

a well-developed theory of analytic functions with values in quasi-Banach spaces is crucial. We

refer to the articles of Kalton [186] and [187] for details. Complex interpolation for sublinear maps

is also possible; see the article of Calderon and Zygmund [47]. Interpolation for analytic families

of operators (Theorem 1.3.7) is due to Stein [331]. The critical Lemma 1.3.8 used in the proof was

previously obtained by Hirschman [154].

The fact that nonatomic measure spaces contain subsets of all possible measures is classical.

An extension of this result to countably additive vector measures with values in finite-dimensional

Banach spaces was obtained by Lyapunov [236]; for a proof of this fact, see Diestel and Uhl [95,

p. 264]. The Aoki–Rolewicz theorem (Exercise 1.4.6) was proved independently by Aoki [7] and

Rolewicz [297]. For a proof of this fact and a variety of its uses in the context of quasi-Banach

spaces we refer to the book of Kalton, Peck, and Roberts [188].

Decreasing rearrangements of functions were introduced by Hardy and Littlewood [146]; the

authors attribute their motivation to understanding cricket averages. The Lp,q spaces were intro-

duced by Lorentz in [232] and in [233]. A general treatment of Lorentz spaces is given in the

article of Hunt [164]. The normability of the spaces Lp,q (which holds exactly when 1 < p ≤ ∞and 1 ≤ q ≤ ∞) can be traced back to general principles obtained by Kolmogorov [199]. The in-

troduction of the function f ∗∗, which was used in Exercise 1.4.3, to explicitly define a norm on

the normable spaces Lp,q is due to Calderon [42]. These spaces appear as intermediate spaces in

the general interpolation theory of Calderon [42] and in that of Lions and Peetre [225]. The latter

was pointed out by Peetre [275]. For a systematic study of the duals of Lorentz spaces we refer to

Cwikel [83] and Cwikel and Fefferman [84], [85]. An extension of the Marcinkiewicz interpolation

theorem to Lorentz spaces was obtained by Hunt [163]. Carro, Raposo, and Soria [57] provide a

comprehensive presentation of the theory of Lorentz spaces in the context of weighted inequali-

ties. For further topics on interpolation one may consult the books of Bennett and Sharpley [24],

Bergh and Lofstrom [25], Sadosky [309], Kislyakov and Kruglyak [194], and Chapter 5 in Stein

and Weiss [348].

Chapter 2

Maximal Functions, Fourier Transform,and Distributions

We have already seen that the convolution of a function with a fixed density is a

smoothing operation that produces a certain average of the function. Averaging is an

important operation in analysis and naturally arises in many situations. The study of

averages of functions is better understood by the introduction of the maximal func-

tion which is defined as the largest average of a function over all balls containing a

fixed point. Maximal functions are used to obtain almost everywhere convergence

for certain integral averages and play an important role in this area, which is called

differentiation theory. Although maximal functions do not preserve qualitative in-

formation about the given functions, they maintain crucial quantitative information,

a fact of great importance in the subject of Fourier analysis.

Another important operation we study in this chapter is the Fourier transform,

the father of all oscillatory integrals. This is as fundamental to Fourier analysis as

marrow is to the human bone. It is a powerful transformation that carries a func-

tion from its spatial domain to its frequency domain. By doing this, it inverts the

function’s localization properties. If applied one more time, then magically repro-

duces the function composed with a reflection. It changes convolution to multipli-

cation, translation to modulation, and expanding dilation to shrinking dilation. Its

decay at infinity encodes information about the local smoothness of the function.

The study of the Fourier transform also motivates the launch of a thorough study

of general oscillatory integrals. We take a quick look at this topic with emphasis on

one-dimensional results.

Distributions suppy a mathematical framework for many operations that do not

exactly qualify to be called functions. These operations found their mathematical

place in the world of functionals applied to smooth functions (called test functions).

These functionals also introduced the correct interpretation for many physical ob-

jects, such as the Dirac delta function. Distributions have become an indispensable

tool in analysis and have enhanced our perspective.



85

86 2 Maximal Functions, Fourier Transform, and Distributions

2.1 Maximal Functions

Given a Lebesgue measurable subset A of Rn, we denote by |A| its Lebesgue mea-

sure. For x ∈Rn and r > 0, we denote by B(x,r) the open ball of radius r centered at

x. We also use the notation aB(x,δ ) = B(x,aδ ), for a > 0, for the ball with the same

center and radius aδ . Given δ > 0 and f a locally integrable function on Rn, let

AvgB(x,δ )

| f |= 1

|B(x,δ )|

∫

B(x,δ )| f (y)|dy

denote the average of | f | over the ball of radius δ centered at x.

2.1.1 The Hardy–Littlewood Maximal Operator

Definition 2.1.1. Let f be a locally integrable function on Rn. The function

M( f )(x) = supδ>0

AvgB(x,δ )

| f |= supδ>0

1

vnδ n

∫

|y|<δ| f (x− y)|dy

is called the centered Hardy–Littlewood maximal function of f .

Obviously we have M( f ) =M(| f |)≥ 0; thus the maximal function is a positive

operator. Information concerning cancellation of the function f is lost by passing

to M( f ). We show later that M( f ) pointwise controls f (i.e., M( f ) ≥ | f | almost

everywhere). Note that M maps L∞ to itself, that is, we have

∥∥M( f )∥∥

L∞≤

∥∥ f∥∥

L∞.

Let us compute the Hardy–Littlewood maximal function of a specific function.

Example 2.1.2. On R, let f be the characteristic function of the interval [a,b]. For

x ∈ (a,b), clearly M( f ) = 1. For x ≥ b, a simple calculation shows that the largest

average of f over all intervals (x−δ ,x+δ ) is obtained when δ = x−a. Similarly,

when x≤ a, the largest average is obtained when δ = b− x. Therefore,

M( f )(x) =

⎧⎪⎨⎪⎩

(b−a)/2|x−b| when x≤ a ,

1 when x ∈ (a,b) ,

(b−a)/2|x−a| when x≥ b .

Observe that M( f ) has a jump at x = a and x = b equal to one-half that of f .

M is a sublinear operator, i.e., it satisfies M( f +g)≤M( f )+M(g) and M(λ f )=|λ |M( f ) for all locally integrable functions f and g and all complex constants λ . It

also has some interesting properties:

2.1 Maximal Functions 87

If f is locally integrable, then by considering the average of f over the ball

B(x, |x|+R), which contains the ball B(0,R), we obtain

M( f )(x)≥∫

B(0,R) | f (y)|dy

vn(|x|+R)n, (2.1.1)

for all x ∈ Rn, where vn is the volume of the unit ball in Rn. An interesting conse-

quence of (2.1.1) is the following: suppose that f = 0 on a set of positive measure

E, then M( f ) is not in L1(Rn). In other words, if f is in L1loc(R

n) and M( f ) is in

L1(Rn), then f = 0 a.e. To see this, integrate (2.1.1) over the ball Rn to deduce that

‖ f χB(0,R)‖L1 = 0 and thus f (x) = 0 for almost all x in the ball B(0,R). Since this is

valid for all R = 1,2,3, . . . , it follows that f = 0 a.e. in Rn.

Another remarkable locality property of M is that if M( f )(x0) = 0 for some x0 in

Rn, then f = 0 a.e. To see we take x = x0 in (2.1.1) to deduce that ‖ f χB(0,R)‖L1 = 0

and as before we have that f = 0 a.e. on every ball centered at the origin, i.e., f = 0

a.e. in Rn.

A related analogue of M( f ) is its uncentered version M( f ), defined as the supre-

mum of all averages of f over all open balls containing a given point.

Definition 2.1.3. The uncentered Hardy–Littlewood maximal function of f ,

M( f )(x) = supδ>0

|y−x|<δ

AvgB(y,δ )

| f | ,

is defined as the supremum of the averages of | f | over all open balls B(y,δ ) that

contain the point x.

Clearly M( f ) ≤M( f ); in other words, M is a larger operator than M. However,

M( f )≤ 2nM( f ) and the boundedness properties of M are identical to those of M.

Example 2.1.4. On R, let f be the characteristic function of the interval I = [a,b].For x ∈ (a,b), clearly M( f )(x) = 1. For x > b, a calculation shows that the largest

average of f over all intervals (y− δ ,y+ δ ) that contain x is obtained when δ =12(x− a) and y = 1

2(x+ a). Similarly, when x < a, the largest average is obtained

when δ = 12(b− x) and y = 1

2(b+ x). We conclude that

M( f )(x) =

⎧⎪⎨⎪⎩

(b−a)/|x−b| when x≤ a ,

1 when x ∈ (a,b) ,

(b−a)/|x−a| when x≥ b .

Observe that M does not have a jump at x = a and x = b and is in fact equal to the

function(1+ dist (x,I)

|I|)−1

.

We are now ready to obtain some basic properties of maximal functions. We need

the following simple covering lemma.

Lemma 2.1.5. Let B1,B2, . . . ,Bk be a finite collection of open balls in Rn. Then

there exists a finite subcollection B j1 , . . . ,B jl of pairwise disjoint balls such that

l

∑r=1

∣∣B jr

∣∣≥ 3−n∣∣

k⋃

i=1

Bi

∣∣ . (2.1.2)

Proof. Let us reindex the balls so that

|B1| ≥ |B2| ≥ · · · ≥ |Bk| .

Let j1 = 1. Having chosen j1, j2, . . . , ji, let ji+1 be the least index s > ji such that⋃im=1 B jm is disjoint from Bs. Since we have a finite number of balls, this process will

terminate, say after l steps. We have now selected pairwise disjoint balls B j1 , . . . ,B jl .

If some Bm was not selected, that is, m /∈ j1, . . . , jl, then Bm must intersect a

selected ball B jr for some jr < m. Then Bm has smaller size than B jr and we must

have Bm 3B jr . This shows that the union of the unselected balls is contained in the

union of the triples of the selected balls. Therefore, the union of all balls is contained

in the union of the triples of the selected balls. Thus

∣∣∣∣k⋃

i=1

Bi

∣∣∣∣≤∣∣∣∣

l⋃

r=1

3B jr

∣∣∣∣≤l

∑r=1

|3B jr |= 3nl

∑r=1

|B jr | ,

and the required conclusion follows.

It was noted earlier that M( f ) and M( f ) never map into L1. However, it is true

that these functions are in L1,∞ when f is in L1. Operators that map L1 to L1,∞ are

said to be weak type (1,1). The centered and uncentered maximal functions M and

M are of weak type (1,1) as shown in the next theorem.

Theorem 2.1.6. The uncentered and centered Hardy–Littlewood maximal operators

M and M map L1(Rn) to L1,∞(Rn) with constant at most 3n and also Lp(Rn) to

Lp(Rn) for 1 α

∣∣≤ 3n

α

∫

M( f )>α| f (y)|dy . (2.1.3)

Proof. We claim that the set Eα = x ∈ Rn : M( f )(x) > α is open. Indeed, for

x ∈ Eα , there is an open ball Bx that contains x such that the average of | f | over Bx

is strictly bigger than α . Then the uncentered maximal function of any other point

in Bx is also bigger than α , and thus Bx is contained in Eα . This proves that Eα is

open.

Let K be a compact subset of Eα . For each x ∈ K there exists an open ball Bx

containing the point x such that

∫

Bx

| f (y)|dy > α|Bx| . (2.1.4)

Observe that Bx ⊂ Eα for all x. By compactness there exists a finite subcover

Bx1, . . . ,Bxk

of K. Using Lemma 2.1.5 we find a subcollection of pairwise disjoint

balls Bx j1, . . . ,Bx jl

such that (2.1.2) holds. Using (2.1.4) and (2.1.2) we obtain

|K| ≤∣∣∣

k⋃

i=1

Bxi

∣∣∣≤ 3nl

∑i=1

|Bx ji| ≤ 3n

α

l

∑i=1

∫

Bx ji

| f (y)|dy≤ 3n

α

∫

Eα

| f (y)|dy ,

since all the balls Bx jiare disjoint and contained in Eα . Taking the supremum over

all compact K ⊆ Eα and using the inner regularity of Lebesgue measure, we deduce

(2.1.3). We have now proved that M maps L1 → L1,∞ with constant 3n. It is a trivial

fact that M maps L∞→ L∞ with constant 1. Since M is well defined and finite a.e.

on L1 + L∞, it is also on Lp(Rn) for 1 < p < ∞. The Marcinkiewicz interpolation

theorem (Theorem 1.3.2) implies that M maps Lp(Rn) to Lp(Rn) for all 1 < p < ∞.

Using Exercise 1.3.3, we obtain the following estimate for the operator norm of M

on Lp(Rn):∥∥M

∥∥Lp→Lp ≤

p3np

p−1. (2.1.5)

Observe that a direct application of Theorem 1.3.2 would give the slightly worse

bound of 2(

pp−1

) 1p 3

np . Finally the boundedness of M follows from that of M.

Remark 2.1.7. The previous proof gives a bound on the operator norm of M on

Lp(Rn) that grows exponentially with the dimension. One may wonder whether this

bound could be improved to a better one that does not grow exponentially in the

dimension n, as n→ ∞. This is not possible; see Exercise 2.1.8.

Example 2.1.8. Let R > 0. Then we have

Rn

(|x|+R)n≤M(χB(0,R))(x)≤

6n Rn

(|x|+R)n. (2.1.6)

The lower estimate in (2.1.6), is an easy consequence of the fact that the ball

B(x, |x|+R) contains the ball B(0,R). For the upper estimate, we first consider the

case where |x| ≤ 2R, when clearly M(χB(0,R))(x) ≤ 1 ≤ 3n Rn

(|x|+R)n . In the case where

|x|> 2R, if the balls B(x,r) and B(0,R) intersect, we must have that r > |x|−R. But

note that |x|−R > 13(|x|+R), since |x|> 2R. We conclude that for |x|> 2R we have

M(χB(0,R))(x)≤ supr>0

|B(x,r)∩B(0,R)||B(x,r)| ≤ sup

r>|x|−R

vnRn

vnrn≤ Rn

(13(|x|+R)

)n

and thus the upper estimate in (2.1.6) holds since M(χB(0,R))≤ 2nM(χB(0,R)). Thus

in both cases the upper estimate in (2.1.6) is valid.

Next we estimate M(M(χB(0,R)))(x). First we write

Rn

(|x|+R)n≤ χB(0,R)+

∞

∑k=0

Rn

(R+2kR)nχB(0,2k+1R)\B(0,2kR) .


Using the upper estimate in (2.1.6) and the sublinearity of M, we obtain

M

(Rn

(| · |+R)n

)(x) ≤M(χB(0,R))(x)+

∞

∑k=0

1

(1+2k)nM(χB(0,2k+1R))(x)

≤ 6n Rn

(|x|+R)n+

∞

∑k=0

1

2nk

6n (2k+1R)n

(|x|+2k+1R)n

≤ Cn log(e+ |x|/R)

(1+ |x|/R)n,

where the last estimate follows by summing separately over k satisfying 2k+1≤ |x|/R

and 2k+1 ≥ |x|/R. Note that the presence of the logarithm does not affect the Lp

boundedness of this function when p > 1.

2.1.2 Control of Other Maximal Operators

We now study some properties of the Hardy–Littlewood maximal function. We

begin with a notational definition that we plan to use throughout this book.

Definition 2.1.9. Given a function g on Rn and ε > 0, we denote by gε the following

function:

gε(x) = ε−ng(ε−1x) . (2.1.7)

As observed in Example 1.2.17, if g is an integrable function with integral equal

to 1, then the family defined by (2.1.7) is an approximate identity. Therefore, convo-

lution with gε is an averaging operation. The Hardy–Littlewood maximal function

M( f ) is obtained as the supremum of the averages of a function f with respect to

the dilates of the kernel k = v−1n χB(0,1) in Rn; here vn is the volume of the unit ball

B(0,1). Indeed, we have

M( f )(x) = supε>0

1

vnεn

∫

Rn| f (x− y)|χB(0,1)

( y

ε

)dy

= supε>0

(| f | ∗ kε)(x) .

Note that the function k = v−1n χB(0,1) has integral equal to 1, and convolving with kε

is an averaging operation.

It turns out that the Hardy–Littlewood maximal function controls the averages of

a function with respect to any radially decreasing L1 function. Recall that a function

f on Rn is called radial if f (x) = f (y) whenever |x| = |y|. Note that a radial func-

tion f on Rn has the form f (x) = ϕ(|x|) for some function ϕ on R+. We have the

following result.

Theorem 2.1.10. Let k ≥ 0 be a function on [0,∞) that is continuous except at a

finite number of points. Suppose that K(x) = k(|x|) is an integrable function on Rn

that satisfies

K(x)≥ K(y), whenever |x| ≤ |y|, (2.1.8)

i.e., k is decreasing. Then the following estimate is true:

supε>0

(| f | ∗Kε)(x)≤∥∥K

∥∥L1M( f )(x) (2.1.9)

for all locally integrable functions f on Rn.

Proof. We prove (2.1.9) when K is radial, satisfies (2.1.8), and is compactly sup-

ported and continuous. When this case is established, select a sequence K j of radial,

compactly supported, continuous functions that increase to K as j→∞. This is pos-

sible, since the function k is continuous except at a finite number of points. If (2.1.9)

holds for each K j, passing to the limit implies that (2.1.9) also holds for K. Next,

we observe that it suffices to prove (2.1.9) for x = 0. When this case is established,

replacing f (t) by f (t + x) implies that (2.1.9) holds for all x.

Let us now fix a radial, continuous, and compactly supported function K with

support in the ball B(0,R), satisfying (2.1.8). Also fix an f ∈ L1loc and take x = 0. Let

e1 be the vector (1,0,0, . . . ,0) on the unit sphere Sn−1. Polar coordinates give

∫

Rn| f (y)|Kε(−y)dy =

∫ ∞

0

∫

Sn−1| f (rθ)|Kε(re1)r

n−1 dθ dr . (2.1.10)

Define functions

F(r) =∫

Sn−1| f (rθ)|dθ ,

G(r) =∫ r

0F(s)sn−1 ds ,

where dθ denotes surface measure on Sn−1. Using these functions, (2.1.10), and

integration by parts, we obtain

∫

Rn| f (y)|Kε(y)dy =

∫ εR

0F(r)rn−1Kε(re1)dr

= G(εR)Kε(εRe1)−G(0)Kε(0)−∫ εR

0G(r)dKε(re1)

=∫ ∞

0G(r)d(−Kε(re1)) , (2.1.11)

where two of the integrals are of Lebesgue–Stieltjes type and we used our assump-

tions that G(0) = 0, Kε(0)<∞, G(εR)<∞, and Kε(εRe1) = 0. Let vn be the volume

of the unit ball in Rn. Since

G(r) =∫ r

0F(s)sn−1 ds =

∫

|y|≤r| f (y)|dy≤M( f )(0)vnrn ,

it follows that the expression in (2.1.11) is dominated by

M( f )(0)vn

∫ ∞

0rnd(−Kε(re1)) = M( f )(0)

∫ ∞

0nvnrn−1Kε(re1)dr

= M( f )(0)∥∥K

∥∥L1 .

Here we used integration by parts and the fact that the surface measure of the unit

sphere Sn−1 is equal to nvn. See Appendix A.3. The theorem is now proved.

Remark 2.1.11. Theorem 2.1.10 can be generalized as follows. If K is an L1 function

on Rn such that |K(x)| ≤ k0(|x|) =K0(x), where k0 is a nonnegative decreasing func-

tion on [0,∞) that is continuous except at a finite number of points, then (2.1.9) holds

with ‖K‖L1 replaced by ‖K0‖L1 . Such a K0 is called a radial decreasing majorant of

K. This observation is formulated as the following corollary.

Corollary 2.1.12. If a function ϕ has an integrable radially decreasing majorant Φ ,

then the estimate

supt>0

|( f ∗ϕt)(x)| ≤∥∥Φ

∥∥L1M( f )(x)

is valid for all locally integrable functions f on Rn.

Example 2.1.13. Let

P(x) =cn

(1+ |x|2) n+12

,

where cn is a constant such that

∫

RnP(x)dx = 1 .

The function P is called the Poisson kernel. We define L1 dilates Pt of the Poisson

kernel P by setting

Pt(x) = t−nP(t−1x)

for t > 0. It is straightforward to verify that when n≥ 2,

d2

dt2Pt +

n

∑j=1

∂ 2j Pt = 0 ,

that is, Pt(x1, . . . ,xn) is a harmonic function of the variables (x1, . . . ,xn, t). Therefore,

for f ∈ Lp(Rn), 1≤ p < ∞, the function

u(x, t) = ( f ∗Pt)(x)

is harmonic in Rn+1+ and converges to f (x) in Lp(dx) as t → 0, since Ptt>0 is an

approximate identity. If we knew that f ∗Pt converged to f a.e. as t → 0, then we

could say that u(x, t) solves the Dirichlet problem


∂ 2t u+

n

∑j=1

∂ 2j u = 0 on Rn+1

+ ,

u(x,0) = f (x) a.e. on Rn.

(2.1.12)

Solving the Dirichlet problem (2.1.12) motivates the study of the almost everywhere

convergence of the expressions f ∗Pt .

Let us now compute the value of the constant cn. Denote by ωn−1 the surface area

of Sn−1. Using polar coordinates, we obtain

1

cn

=∫

Rn

dx

(1+ |x|2) n+12

= ωn−1

∫ ∞

0

rn−1

(1+ r2)n+1

2

dr

= ωn−1

∫ π/2

0(sinϕ)n−1 dϕ (r = tanϕ)

=2π

n2

Γ ( n2)

1

2

Γ ( n2)Γ ( 1

2)

Γ ( n+12)

=π

n+12

Γ ( n+12),

where we used the formula for ωn−1 in Appendix A.3 and an identity in Appendix

A.4. We conclude that

cn =Γ ( n+1

2)

πn+1

2

and that the Poisson kernel on Rn is given by

P(x) =Γ ( n+1

2)

πn+1

2

1

(1+ |x|2) n+12

. (2.1.13)

Theorem 2.1.10 implies that the solution of the Dirichlet problem (2.1.12) is point-

wise bounded by the Hardy–Littlewood maximal function of f .

2.1.3 Applications to Differentiation Theory

We continue this section by obtaining some applications of the boundedness of the

Hardy–Littlewood maximal function in differentiation theory.

We now show that the weak type (1,1) property of the Hardy–Littlewood max-

imal function implies almost everywhere convergence for a variety of families of

functions. We deduce this from the more general fact that a certain weak type prop-

erty for the supremum of a family of linear operators implies almost everywhere

convergence.

Here is our setup. Let (X ,µ), (Y,ν) be measure spaces and let 0 0 there exists a g ∈ D such that ‖ f −g‖Lp < δ . Suppose that for

every ε > 0, Tε is a linear operator that maps Lp(X ,µ) into a subspace of measurable

functions, which are defined everywhere on Y . For y∈Y , define a sublinear operator

T∗( f )(y) = supε>0

|Tε( f )(y)| (2.1.14)

and assume that T∗( f ) is ν- measurable for any f ∈ Lp(X ,µ). We have the following.

Theorem 2.1.14. Let 0 0 and all f ∈ Lp(X) we have

∥∥T∗( f )∥∥

Lq,∞ ≤ B∥∥ f

∥∥Lp (2.1.15)

and that for all f ∈ D,

limε→0

Tε( f ) = T ( f ) (2.1.16)

exists and is finite ν-a.e. (and defines a linear operator on D). Then for all func-

tions f in Lp(X ,µ) the limit (2.1.16) exists and is finite ν-a.e., and defines a linear

operator T on Lp(X) (uniquely extending T defined on D) that satisfies

∥∥T ( f )∥∥

Lq,∞ ≤ B∥∥ f

∥∥Lp (2.1.17)

for all functions f in Lp(X).

Proof. Given f in Lp, we define the oscillation of f :

O f (y) = limsupε→0

limsupθ→0

|Tε( f )(y)−Tθ ( f )(y)| .

We would like to show that for all f ∈ Lp and δ > 0,

ν(y ∈ Y : O f (y)> δ) = 0 . (2.1.18)

Once (2.1.18) is established, given f ∈ Lp(X), we obtain that O f (y) = 0 for ν-almost

all y, which implies that Tε( f )(y) is Cauchy for ν-almost all y, and it therefore

converges ν-a.e. to some T ( f )(y) as ε → 0. The operator T defined this way on

Lp(X) is linear and extends T defined on D.

To approximate O f we use density. Given η > 0, find a function g ∈ D such that

‖ f − g‖Lp < η . Since Tε(g)→ T (g) ν-a.e, it follows that Og = 0 ν-a.e. Using this

fact and the linearity of the Tε ’s, we conclude that

O f (y)≤ Og(y)+O f−g(y) = O f−g(y) ν-a.e.

Now for any δ > 0 we have

ν(y ∈ Y : O f (y)> δ) ≤ ν(y ∈ Y : O f−g(y)> δ)≤ ν(y ∈ Y : 2T∗( f −g)(y)> δ)≤

(2B

∥∥ f −g∥∥

Lp/δ)q

≤ (2Bη/δ )q .

Letting η → 0, we deduce (2.1.18). We conclude that Tε( f ) is a Cauchy sequence,

and hence it converges ν-a.e. to some T ( f ). Since |T ( f )| ≤ |T∗( f )|, the conclusion

(2.1.17) of the theorem follows easily.

We now derive some applications. First we return to the issue of almost every-

where convergence of the expressions f ∗Py, where P is the Poisson kernel.

Example 2.1.15. Fix 1≤ p < ∞ and f ∈ Lp(Rn). Let

P(x) =Γ ( n+1

2)

πn+1

2

1

(1+ |x|2) n+12

be the Poisson kernel on Rn and let Pε(x) = ε−nP(ε−1x). We deduce from the previ-

ous theorem that the family f ∗Pε converges to f a.e. Let D be the set of all contin-

uous functions with compact support on Rn. Since the family (Pε)ε>0 is an approx-

imate identity, Theorem 1.2.19 (2) implies that for f in D we have that f ∗Pε → f

uniformly on compact subsets of Rn and hence pointwise everywhere. In view of

Theorem 2.1.10, the supremum of the family of linear operators Tε( f ) = f ∗Pε is

controlled by the Hardy–Littlewood maximal function, and thus it maps Lp to Lp,∞

for 1≤ p<∞. Theorem 2.1.14 now gives that f ∗Pε converges to f a.e. for all f ∈ Lp.

Here is another application of Theorem 2.1.14. Exercise 2.1.10 contains other ap-

plications.

Corollary 2.1.16. (Lebesgue’s differentiation theorem) For any locally integrable

function f on Rn we have

limr→0

1

|B(x,r)|

∫

B(x,r)f (y)dy = f (x) (2.1.19)

for almost all x in Rn. Consequently we have | f | ≤ M( f ) a.e. There is also an

analogous statement to (2.1.19) in which balls are replaced by cubes centered at x.

Precisely, for any locally integrable function f on Rn we have

limr→0

1

(2r)n

∫

x+[−r,r]nf (y)dy = f (x) (2.1.20)

for almost all x in Rn.

Proof. Since Rn is the union of the balls B(0,N) for N = 1,2,3 . . . , it suffices to

prove the required conclusion for almost all x inside a fixed ball B(0,N). Given a

locally integrable function f on Rn, consider the function fN = f χB(0,N+1). Then

fN lies in L1(Rn). Let Tε be the operator given with convolution with kε , where

k = v−1n χB(0,1) and 0 < ε < 1. We know that the corresponding maximal operator T∗

is controlled by the centered Hardy–Littlewood maximal function M, which maps L1

to L1,∞. It is straightforward to verify that (2.1.19) holds for all continuous functions

f with compact support. Since this set of functions is dense in L1, and T∗ maps L1

to L1,∞, Theorem 2.1.14 implies that (2.1.19) holds for all integrable functions on Rn,

in particular for fN . But for 0 < ε < 1 and x ∈ B(0,N) we have f χB(x,ε) = fNχB(x,ε),

so it follows that

limε→0

1

|B(x,ε)|

∫

B(x,ε)f (y)dy = lim

ε→0

1

|B(x,ε)|

∫

B(x,ε)fN(y)dy = fN(x)

for almost all x ∈Rn, in particular for almost all x in B(0,N). But on this set fN = f ,

so the required conclusion follows. The assertion that | f | ≤M( f ) a.e. is an easy

consequence of (2.1.19) when the limit is replaced by a supremum.

Finally, with minor modifications, the proof can be adjusted to work for cubes in

place of balls. To prove (2.1.20), for f ∈ L1loc(R

n) we introduce the maximal operator

Mc( f )(x) = supr>0

1

(2r)n

∫

x+[−r,r]n| f (y)|dy .

Then Exercise 2.1.3 yields that Mc maps L1(Rn) to weak L1(Rn) and the preceding

proof with Mc in place of M yields (2.1.20).

The following corollaries were inspired by Example 2.1.15.

Corollary 2.1.17. (Differentiation theorem for approximate identities) Let K be an

L1 function on Rn with integral 1 that has a continuous integrable radially decreas-

ing majorant. Then f ∗Kε → f a.e. as ε → 0 for all f ∈ Lp(Rn), 1≤ p < ∞.

Proof. It follows from Example 1.2.17 that Kε is an approximate identity. Theorem

1.2.19 now implies that f ∗Kε → f uniformly on compact sets when f is continuous.

Let D be the space of all continuous functions with compact support. Then f ∗Kε →f a.e. for f ∈ D. It follows from Corollary 2.1.12 that T∗( f ) = supε>0 | f ∗Kε | maps

Lp to Lp,∞ for 1 ≤ p < ∞. Using Theorem 2.1.14, we conclude the proof of the

corollary.

Remark 2.1.18. Fix f ∈ Lp(Rn) for some 1 ≤ p < ∞. Theorem 1.2.19 implies that

f ∗Kε converges to f in Lp and hence some subsequence f ∗Kεn of f ∗Kε converges

to f a.e. as n→ ∞, (εn → 0). Compare this result with Corollary 2.1.17, which gives

a.e. convergence for the whole family f ∗Kε as ε → 0.

Corollary 2.1.19. (Differentiation theorem for multiples of approximate identi-

ties) Let K be a function on Rn that has an integrable radially decreasing majorant.

Let a =∫

Rn K(x)dx. Then for all f ∈ Lp(Rn) and 1 ≤ p < ∞, ( f ∗Kε)(x)→ a f (x)for almost all x ∈ Rn as ε → 0.

Proof. Use Theorem 1.2.21 instead of Theorem 1.2.19 in the proof of Corollary

2.1.17.

The following application of the Lebesgue differentiation theorem uses a simple

stopping-time argument. This is the sort of argument in which a selection procedure

stops when it is exhausted at a certain scale and is then repeated at the next scale. A

certain refinement of the following proposition is of fundamental importance in the

study of singular integrals given in Chapter 4.

Proposition 2.1.20. Given a nonnegative integrable function f on Rn and α > 0,

there exists a collection of disjoint (possibly empty) open cubes Q j such that for

almost all x ∈(⋃

j Q j

)cwe have f (x)≤ α and

α <1

|Q j|

∫

Q j

f (t)dt ≤ 2nα . (2.1.21)

Proof. The proof provides an excellent paradigm of a stopping-time argument. Start

by decomposing Rn as a union of cubes of equal size, whose interiors are disjoint,

and whose diameter is so large that |Q|−1∫

Q f (x)dx ≤ α for every Q in this mesh.

This is possible since f is integrable and |Q|−1∫

Q f (x)dx→ 0 as |Q| → ∞. Call the

union of these cubes E0.

Divide each cube in the mesh into 2n congruent cubes by bisecting each of the

sides. Call the new collection of cubes E1. Select a cube Q in E1 if

1

|Q|

∫

Qf (x)dx > α (2.1.22)

and call the set of all selected cubes S1. Now subdivide each cube in E1 \S1 into

2n congruent cubes by bisecting each of the sides as before. Call this new collection

of cubes E2. Repeat the same procedure and select a family of cubes S2 that satisfy

(2.1.22). Continue this way ad infinitum and call the cubes in⋃∞

m=1 Sm “selected.”

If Q was selected, then there exists Q1 in Em−1 containing Q that was not selected at

the (m−1)th step for some m≥ 1. Therefore,

α <1

|Q|

∫

Qf (x)dx≤ 2n 1

|Q1|

∫

Q1

f (x)dx≤ 2nα .

Now call F the closure of the complement of the union of all selected cubes. If

x ∈ F , then there exists a sequence of cubes containing x whose diameter shrinks

down to zero such that the average of f over these cubes is less than or equal to α .

By Corollary 2.1.16, it follows that f (x) ≤ α almost everywhere in F . This proves

the proposition.

In the proof of Proposition 2.1.20 it was not crucial to assume that f was defined

on all Rn, but only on a cube. We now give a local version of this result.

Corollary 2.1.21. Let f ≥ 0 be an integrable function over a cube Q in Rn and let

α ≥ 1|Q|

∫Q f dx. Then there exist disjoint (possibly empty) open subcubes Q j of Q

such that for almost all x ∈ Q\⋃ j Q j we have f ≤ α and (2.1.21) holds for all Q j.

Proof. The proof easily follows by a simple modification of Proposition 2.1.20 in

which Rn is replaced by the fixed cube Q. To apply Corollary 2.1.16, we extend f to

be zero outside the cube Q.

See Exercise 2.1.4 for an application of Proposition 2.1.20 involving maximal

functions.

Exercises

2.1.1. A positive Borel measure µ on Rn is called inner regular if for any open

subset U of Rn we have µ(U) = supµ(K) : K U, K compact and µ is called

locally finite if µ(B)< ∞ for all balls B.

(a) Let µ be a positive inner regular locally finite measure on Rn that satisfies the

following doubling condition: There exists a constant D(µ) > 0 such that for all

x ∈ Rn and r > 0 we have

µ(3B(x,r))≤ D(µ)µ(B(x,r)).

For f ∈ L1loc(R

n,µ) define the uncentered maximal function Mµ( f ) with respect to

µ by

Mµ( f )(x) = supr>0

supz: |z−x|<r

µ(B(z,r)) =0

1

µ(B(z,r))

∫

B(z,r)f (y)dµ(y) .

Show that Mµ maps L1(Rn,µ) to L1,∞(Rn,µ) with constant at most D(µ) and

Lp(Rn,µ) to itself with constant at most 2(

pp−1

) 1p D(µ)

1p .

(b) Obtain as a consequence a differentiation theorem analogous to Corollary 2.1.16.[Hint: Part (a): For f ∈ L1(Rn,µ) show that the set Eα = Mµ( f ) > α is open.

Then use the argument of the proof of Theorem 2.1.6 and the inner regularity of µ .]

2.1.2. On R consider the maximal function Mµ of Exercise 2.1.1.

(a) (W. H. Young) Prove the following covering lemma. Given a finite set F of open

intervals in R, prove that there exist two subfamilies each consisting of pairwise dis-

joint intervals such that the union of the intervals in the original family is equal to the

union of the intervals of both subfamilies. Use this result to show that the maximal

function Mµ of Exercise 2.1.1 maps L1(µ)→ L1,∞(µ) with constant at most 2.

(b) ([134]) Prove that for any σ -finite positive measure µ on R, α > 0, and f ∈L1

loc(R,µ) we have

1

α

∫

A| f |dµ−µ(A)≤ 1

α

∫

| f |>α| f |dµ−µ(| f |> α) .

Use this result and part (a) to prove that for all α > 0 and all locally integrable f we

have

µ(| f |> α)+µ(Mµ( f )> α)≤ 1

α

∫

| f |>α| f |dµ+

1

α

∫

Mµ ( f )>α| f |dµ

and note that equality is obtained when α = 1 and f (x) = |x|−1/p.

(c) Conclude that Mµ maps Lp(µ) to Lp(µ), 1 < p < ∞, with bound at most the

unique positive solution Ap of the equation

(p−1)xp− pxp−1−1 = 0 .

(d) ([136]) If µ is the Lebesgue measure show that for 1 < p < ∞ we have

∥∥M∥∥

Lp→Lp = Ap ,

where Ap is the unique positive solution of the equation in part (c).[Hint: Part (a): Select a subset G of F with minimal cardinality such that

⋃J∈G J =⋃

I∈F I. Part (d): One direction follows from part (c). Conversely, M(|x|−1/p)(1) =p

p−1γ1/p′+1γ+1

, where γ is the unique positive solution of the equationp

p−1γ1/p′+1γ+1

=

γ−1/p. Conclude that M(|x|−1/p)(1) = Ap and that M(|x|−1/p) = Ap|x|−1/p. Since

this function is not in Lp, consider the family fε(x) = |x|−1/p min(|x|−ε , |x|ε), ε > 0,

and show that M( fε)(x)≥ (1+ γ1p′+ε)(1+ γ)−1( 1

p′ + ε)−1 fε(x) for 0 < ε < p′.]

2.1.3. Define the centered Hardy–Littlewood maximal function Mc and the uncen-

tered Hardy–Littlewood maximal function Mc using cubes with sides parallel to the

axes instead of balls in Rn. Prove that

1≤ M( f )

M( f )≤ 2n ,

1

nn2

2n

vn

≤ M( f )

Mc( f )≤ 2n

vn

,1

nn2

2n

vn

≤ M( f )

Mc( f )≤ 2n

vn

,

where vn is the volume of the unit ball in Rn. Conclude that Mc and Mc are weak

type (1,1) and they map Lp(Rn) to Lp(Rn) for 1 < p≤ ∞.

2.1.4. (a) Prove the estimate:

|x ∈ Rn : M( f )(x)> 2α| ≤ 3n

α

∫

| f |>α| f (y)|dy

and conclude that M maps Lp to Lp,∞ with norm at most 2 ·3n/p for 1≤ p < ∞.

(b) Deduce that if f log+(2| f |) is integrable over a ball B, then M( f ) is integrable

over the same ball B.

(c) ([375], [336]) Apply Proposition 2.1.20 to | f | and α > 0 and Exercise 2.1.3 to

show that with cn = 2n(nn/2vn)−1 we have

|x ∈ Rn : M( f )(x)> cnα| ≥2−n

α

∫

| f |>α| f (y)|dy .

(d) Suppose that f is integrable and supported in a ball B(0,ρ). Show that for x in

B(0,2ρ)\B(0,ρ) we have M( f )(x)≤M( f )(ρ2|x|−2x). Conclude that

∫

B(0,2ρ)M( f )dx≤ (4n +1)

∫

B(0,ρ)M( f )dx

and from this deduce a similar inequality for M( f ).(e) Suppose that f is integrable and supported in a ball B and that M( f ) is integrable

over B. Let λ0 = 2n|B|−1‖ f‖L1 . Use part (b) to prove that f log+(λ−10 cn | f |) is inte-

grable over B.[Hint: Part (a): Write f = f χ| f |>α + f χ| f |≤α . Part (b): Show that M( f χE) is inte-

grable over B, where E = | f | ≥ 1/2. Part (c): Use Proposition 2.1.20. Part (d): Let

x′ = ρ2|x|−2x for some ρ < |x|< 2ρ . Show that for R > |x|−ρ , we have that

∫

B(x,R)| f (z)|dz≤

∫

B(x′,R)| f (z)|dz

by showing that B(x,R)∩B(0,ρ)⊂B(x′,R). Part (e): For x /∈ 2B we have M( f )(x)≤λ0, hence

∫2B M( f )(x)dx≥ ∫ ∞

λ0|x ∈ 2B : M( f )(x)> α|dα .

]

2.1.5. (A. Kolmogorov) Let S be a sublinear operator that maps L1(Rn) to L1,∞(Rn)with norm B. Suppose that f ∈ L1(Rn). Prove that for any set A of finite Lebesgue

measure and for all 0 < q < 1 we have

∫

A|S( f )(x)|q dx≤ (1−q)−1Bq|A|1−q

∥∥ f∥∥q

L1 ,

and in particular, for the Hardy–Littlewood maximal operator,

∫

AM( f )(x)q dx≤ (1−q)−13nq|A|1−q

∥∥ f∥∥q

L1 .

[Hint: Use the identity

∫

A|S( f )(x)|q dx =

∫ ∞

0qαq−1|x∈A : S( f )(x)>α|dα

and estimate the last measure by min(|A|, Bα ‖ f‖L1).

]

2.1.6. Let Ms( f )(x) be the supremum of the averages of | f | over all rectangles with

sides parallel to the axes containing x. The operator Ms is called the strong maximal

function.

(a) Prove that Ms maps Lp(Rn) to itself.

(b) Show that the operator norm of Ms is Anp, where Ap is as in Exercise 2.1.2 (c).

(c) Prove that Ms is not weak type (1,1).

2.1.7. Prove that if

|ϕ(x1, . . . ,xn)| ≤ A(1+ |x1|)−1−ε · · ·(1+ |xn|)−1−ε

for some A,ε > 0, and ϕt1,...,tn(x) = t−11 · · · t−1

n ϕ(t−11 x1, . . . , t

−1n xn), then the maximal

operator

f → supt1,...,tn>0

| f ∗ϕt1,...,tn |

is pointwise controlled by the strong maximal function.

2.1.8. Prove that for any fixed 1 < p < ∞, the operator norm of M on Lp(Rn) tends

to infinity as n→ ∞.[Hint: Let f0 be the characteristic function of the unit ball in Rn. Consider the aver-

ages |Bx|−1∫

Bxf0 dy, where Bx = B

(12(|x|− |x|−1) x

|x| ,12(|x|+ |x|−1)

)for |x|> 1.

]

2.1.9. (a) In R2 let M0( f )(x) be the maximal function obtained by taking the supre-

mum of the averages of | f | over all rectangles (of arbitrary orientation) containing

x. Prove that M0 is not bounded on Lp(Rn) for p ≤ 2 and conclude that M0 is not

weak type (1,1).(b) Let M00( f )(x) be the maximal function obtained by taking the supremum of the

averages of | f | over all rectangles in R2 of arbitrary orientation but fixed eccentricity

containing x. (The eccentricity of a rectangle is the ratio of its longer side to its

shorter side.) Using a covering lemma, show that M00 is weak type (1,1) with a

bound proportional to the square of the eccentricity.

(c) On Rn define a maximal function by taking the supremum of the averages

of | f | over all products of intervals I1 × ·· · × In containing a point x with |I2| =a2|I1|, . . . , |In| = an|I1| and a2, . . . ,an > 0 fixed. Show that this maximal function is

of weak type (1,1) with bound independent of the numbers a2, . . . ,an.[Hint: Part (b): Let b be the eccentricity. If two rectangles with the same eccentricity

intersect, then the smaller one is contained in the bigger one scaled 4b times. Then

use an argument similar to that in Lemma 2.1.5.]

2.1.10. (a) Let 0 < p,q < ∞ and let X ,Y be measure spaces. Suppose that Tε are

maps from Lp(X) to Lq,∞(Y ) satisfy |Tε( f +g)| ≤ K(|Tε( f )|+ |Tε(g)|) for all ε > 0

and all f ,g∈ Lp(X), and also limε→0 Tε( f ) = 0 a.e. for all f in some dense subspace

D of Lp(X). Assume furthermore that the maximal operator T∗( f ) = supε>0 |Tε( f )|maps Lp(X) to Lq,∞(Y ). Prove that limε→0 Tε( f ) = 0 a.e. for all f in Lp(X).(b) Use the result in part (a) to prove the following version of the Lebesgue differ-

entiation theorem: Let f ∈ Lp(Rn) for some 0 < p < ∞. Then for almost all x ∈ Rn

we have

lim|B|→0B∋x

1

|B|

∫

B|g(y)−g(x)|p dy = 0 ,

where the limit is taken over all open balls B containing x and shrinking to x.(c) Conclude that for any f in L1

loc(Rn) and for almost all x ∈ Rn we have

lim|B|→0B∋x

1

|B|

∫

Bf (y)dy = f (x) ,

where the limit is taken over all open balls B containing x and shrinking to x.[Hint: (a) Define an oscillation O f (y) = limsupε→0 |Tε( f )(y)|. For all f in Lp(X)

and g ∈ D we have that O f (y) ≤ KO f−g(y). Then use the argument in the proof of

Theorem 2.1.14. (b) Apply part (a) with

Tε( f )(x) = supB(z,ε)∋x

(1

|B(z,ε)|

∫

B(z,ε)| f (y)− f (x)|p dy

)1/p

,

observing that T∗( f ) = supε>0 Tε( f ) ≤ max(1,21−p

p )(| f |+M(| f |p)

1p). (c) Follows

from part (b) with p = 1. Note that part (b) can be proved without part (a) but using

part (c) as follows: for every rational number a there is a set Ea of Lebesgue measure

zero such that for x∈Rn \Ea we have limB∋x,|B|→01|B|

∫B |g(y)−a|p dy = |g(x)−a|p,

since the function y → | f (y)−a|p is in L1loc(R

n). By considering an enumeration of

the rationals, find a set of measure zero E such for x /∈ E the preceding limit exists

for all rationals a and by continuity for all real numbers a, in particular for a= g(x).]

2.1.11. Let f be in L1(R). Define the right maximal function MR( f ) and the left

maximal function ML( f ) as follows:

ML( f )(x) = supr>0

1

r

∫ x

x−r| f (t)|dt ,

MR( f )(x) = supr>0

1

r

∫ x+r

x| f (t)|dt .

(a) Show that for all α > 0 and f ∈ L1(R) we have

|x ∈ R : ML( f )(x)> α| = 1

α

∫

ML( f )>α| f (t)|dt ,

|x ∈ R : MR( f )(x)> α| = 1

α

∫

MR( f )>α| f (t)|dt .

(b) Extend the definition of ML( f ) and MR( f ) for f ∈ Lp(R) for 1 ≤ p ≤ ∞. Show

that ML and MR map Lp to Lp with norm at most p/(p−1) for all p with 1 < p <∞.

(c) Construct examples to show that the operator norms of ML and MR on Lp(R) are

exactly p/(p−1) for 1 < p < ∞.

(d) Prove that M = max(MR,ML).(e) Let N = min(MR,ML). Obtain the following consequence of part (a),

∫

RM( f )p +N( f )p dx =

p

p−1

∫

R| f |

(M( f )p−1 +N( f )p−1

)dx ,

(f) Use part (e) to prove that

(p−1)∥∥M( f )

∥∥p

Lp − p∥∥ f

∥∥Lp

∥∥M( f )∥∥p−1

Lp −∥∥ f

∥∥p

Lp ≤ 0 .

[Hint: (a) Write the set Eα = MR( f ) > α as a union of open intervals (a j,b j).

For each x in (a j,b j), let Nx =

s ∈R :∫ s

x | f |> α(s−x)∩ (x,b j]. Show that Nx is

nonempty and that supNx = b j for every x ∈ (a j,b j). Conclude that∫ b j

a j| f (t)|dt ≥

α(b j− a j), which implies that each a j is finite. For the reverse inequality use that

a j /∈ Eα . Part (d) is due to K. L. Phillips. (e) First obtain a version of the equality

with MR in the place of M and ML in the place of N. Then use that M( f )q+N( f )q =ML( f )q +MR( f )q for all q. (f) Use that | f |N( f )p−1 ≤ 1

p| f |p + 1

p′N( f )p. This alter-

native proof of the result in Exercise 2.1.2(c) was suggested by J. Duoandikoetxea.]

2.1.12. A cube Q = [a12k,(a1 + 1)2k)× ·· · × [an2k,(an + 1)2k) on Rn is called

dyadic if k, a1, . . . ,an ∈ Z. Observe that either two dyadic cubes are disjoint or one

contains the other. Define the dyadic maximal function

Md( f )(x) = supQ∋x

1

|Q|

∫

Qf (y)dy ,

where the supremum is taken over all dyadic cubes Q containing x.

(a) Prove that Md maps L1 to L1,∞ with constant at most one. Presicely, show that for

all α > 0 and f ∈ L1(Rn) we have

|x ∈ Rn : Md( f )(x)> α| ≤ 1

α

∫

Md( f )>αf (t)dt .

(b) Conclude that Md maps Lp(Rn) to itself with constant at most p/(p−1).

2.1.13. Observe that the proof of Theorem 2.1.6 yields the estimate

λ |M( f )> λ|1p ≤ 3n|M( f )> λ|−1+ 1

p

∫

M( f )>λ| f (y)|dy

for λ > 0 and f locally integrable. Use the result of Exercise 1.1.12(a) to prove that

the Hardy–Littlewood maximal operator M maps the space Lp,∞(Rn) to itself for

1 < p < ∞.


2.1.14. Let K(x) = (1+ |x|)−n−δ be defined on Rn. Prove that there exists a constant

Cn,δ such that for all ε0 > 0 we have the estimate

supε>ε0

(| f | ∗Kε)(x)≤Cn,δ supε>ε0

1

εn

∫

|y−x|≤ε| f (y)|dy ,

for all f locally integrable on Rn.[Hint: Apply only a minor modification to the proof of Theorem 2.1.10.

]

2.2 The Schwartz Class and the Fourier Transform

In this section we introduce the single most important tool in harmonic analysis, the

Fourier transform. It is often the case that the Fourier transform is introduced as an

operation on L1 functions. In this exposition we first define the Fourier transform

on a smaller class, the space of Schwartz functions, which turns out to be a very

natural environment. Once the basic properties of the Fourier transform are derived,

we extend its definition to other spaces of functions.

We begin with some preliminaries. Given x = (x1, . . . ,xn) ∈ Rn, we set |x| =(x2

1 + · · ·+ x2n)

1/2. The partial derivative of a function f on Rn with respect to the

jth variable x j is denoted by ∂ j f while the mth partial derivative with respect to

the jth variable is denoted by ∂mj f . The gradient of a function f is the vector ∇ f =

(∂1 f , . . . ,∂n f ). A multi-index α is an ordered n-tuple of nonnegative integers. For

a multi-index α = (α1, . . . ,αn), ∂α f denotes the derivative ∂α1

1 · · ·∂αnn f . If α =

(α1, . . . ,αn) is a multi-index, |α|=α1+ · · ·+αn denotes its size and α!=α1! · · ·αn!

denotes the product of the factorials of its entries. The number |α| indicates the total

order of differentiation of ∂α f . The space of functions in Rn all of whose derivatives

of order at most N ∈ Z+ are continuous is denoted by C N(Rn) and the space of all

infinitely differentiable functions on Rn by C ∞(Rn). The space of C ∞ functions with

compact support on Rn is denoted by C ∞0 (Rn). This space is nonempty; see Exercise

2.2.1(a).

For x ∈ Rn and α = (α1, . . . ,αn) a multi-index, we set xα = xα11 · · ·xαn

n . Multi-

indices will be denoted by the letters α,β ,γ ,δ , .... It is a simple fact to verify that

|xα | ≤ cn,α |x||α | , (2.2.1)

for some constant that depends on the dimension n and on α . In fact, cn,α is

the maximum of the continuous function (x1, . . . ,xn) → |xα11 · · ·xαn

n | on the sphere

Sn−1 = x ∈ Rn : |x| = 1. The converse inequality in (2.2.1) fails. However, the

following substitute of the converse of (2.2.1) is of great use: for k ∈ Z+ we have

|x|k ≤Cn,k ∑|β |=k

|xβ | (2.2.2)

for all x ∈Rn \0. To prove (2.2.2), take 1/Cn,k to be the minimum of the function

x → ∑|β |=k

|xβ |

2.2 The Schwartz Class and the Fourier Transform 105

on Sn−1; this minimum is positive since this function has no zeros on Sn−1. A related

inequality is

(1+ |x|)k ≤ 2k(1+Cn,k) ∑|β |≤k

|xβ | . (2.2.3)

This follows from (2.2.2) for |x| ≥ 1, while for |x|< 1 we note that the sum in (2.2.3)

is at least one since |x(0,...,0)|= 1.

We end the preliminaries by noting the validity of the one-dimensional Leibniz

ruledm

dtm( f g) =

m

∑k=0

(m

k

)dk f

dtk

dm−kg

dtm−k, (2.2.4)

for all C m functions f ,g on R, and its multidimensional analogue

∂α( f g) = ∑β≤α

(α1

β1

)· · ·

(αn

βn

)(∂ β f )(∂α−βg) , (2.2.5)

for f ,g in C |α |(Rn) for some multi-index α , where the notation β ≤ α in (2.2.5)

means that β ranges over all multi-indices satisfying 0≤ β j ≤ α j for all 1≤ j ≤ n.

We observe that identity (2.2.5) is easily deduced by repeated application of (2.2.4),

which in turn is obtained by induction.

2.2.1 The Class of Schwartz Functions

We now introduce the class of Schwartz functions on Rn. Roughly speaking, a func-

tion is Schwartz if it is smooth and all of its derivatives decay faster than the recip-

rocal of any polynomial at infinity. More precisely, we give the following definition.

Definition 2.2.1. A C ∞ complex-valued function f on Rn is called a Schwartz func-

tion if for every pair of multi-indices α and β there exists a positive constant Cα ,β

such that

ρα ,β ( f ) = supx∈Rn

|xα∂ β f (x)|=Cα ,β < ∞. (2.2.6)

The quantities ρα ,β ( f ) are called the Schwartz seminorms of f . The set of all

Schwartz functions on Rn is denoted by S (Rn).

Example 2.2.2. The function e−|x|2

is in S (Rn) but e−|x| is not, since it fails to be

differentiable at the origin. The C ∞ function g(x) = (1+ |x|4)−a, a > 0, is not in S

since it decays only like the reciprocal of a fixed polynomial at infinity. The set of

all smooth functions with compact support, C ∞0 (Rn), is contained in S (Rn).

Remark 2.2.3. If f1 is in S (Rn) and f2 is in S (Rm), then the function of m+ n

variables f1(x1, . . . ,xn) f2(xn+1, . . . ,xn+m) is in S (Rn+m). If f is in S (Rn) and P(x)is a polynomial of n variables, then P(x) f (x) is also in S (Rn). If α is a multi-index

and f is in S (Rn), then ∂α f is in S (Rn). Also note that

f ∈S (Rn) ⇐⇒ supx∈Rn

|∂α(xβ f (x))|< ∞ for all multi-indices α , β .

Remark 2.2.4. The following alternative characterization of Schwartz functions is

very useful. A C ∞ function f is in S (Rn) if and only if for all positive integers N

and all multi-indices α there exists a positive constant Cα ,N such that

|(∂α f )(x)| ≤Cα ,N(1+ |x|)−N . (2.2.7)

The simple proofs are omitted. We now discuss convergence in S (Rn).

Definition 2.2.5. Let fk, f be in S (Rn) for k = 1,2, . . . . We say that the sequence

fk converges to f in S (Rn) if for all multi-indices α and β we have

ρα ,β ( fk− f ) = supx∈Rn

|xα(∂ β ( fk− f ))(x)| → 0 as k→ ∞.

For instance, for any fixed x0 ∈ Rn, f (x+ x0/k)→ f (x) in S (Rn) for any f in

S (Rn) as k→ ∞.

This notion of convergence is compatible with a topology on S (Rn) under which

the operations ( f ,g) → f + g, (a, f )→ a f , and f → ∂α f are continuous for all

complex scalars a and multi-indices α ( f ,g ∈ S (Rn)). A subbasis for open sets

containing 0 in this topology is

f ∈S : ρα ,β ( f )< r ,

for all α , β multi-indices and all r ∈Q+. Observe the following: If ρα ,β ( f ) = 0, then

f = 0. This means that S (Rn) is a locally convex topological vector space equipped

with the family of seminorms ρα ,β that separate points. We refer to Reed and Simon

[286] for the pertinent definitions. Since the origin in S (Rn) has a countable base,

this space is metrizable. In fact, the following is a metric on S (Rn):

d( f ,g) =∞

∑j=1

2− j ρ j( f −g)

1+ρ j( f −g),

where ρ j is an enumeration of all the seminorms ρα ,β , α and β multi-indices. One

may easily verify that S is complete with respect to the metric d. Indeed, a Cauchy

sequence h j j in S would have to be Cauchy in L∞ and therefore it would con-

verge uniformly to some function h. The same is true for the sequences ∂ βh j j

and xαh j(x) j, and the limits of these sequences can be shown to be the functions

∂ βh and xαh(x), respectively. It follows that the sequence h j converges to h in S .

Therefore, S (Rn) is a Frechet space (complete metrizable locally convex space).

We note that convergence in S is stronger than convergence in all Lp. We have

the following.

Proposition 2.2.6. Let f , fk, k = 1,2,3, . . . , be in S (Rn). If fk → f in S then

fk → f in Lp for all 0 < p≤ ∞. Moreover, there exists a Cp,n > 0 such that

∥∥∂ β f∥∥

Lp ≤Cp,n ∑|α |≤[ n+1

p ]+1

ρα ,β ( f ) (2.2.8)

for all f for which the right-hand side is finite.

Proof. Observe that when p < ∞ we have

∥∥∂ β f∥∥

Lp ≤[∫

|x|≤1|∂ β f (x)|p dx+

∫

|x|≥1|x|n+1|∂ β f (x)|p|x|−(n+1) dx

]1/p

≤[

vn

∥∥∂ β f∥∥p

L∞+

(sup|x|≥1

|x|n+1|∂ β f (x)|p)∫

|x|≥1|x|−(n+1) dx

]1/p

≤Cp,n

(∥∥∂ β f∥∥

L∞+ sup|x|≥1

(|x|[n+1

p ]+1|∂ β f (x)|)).

The preceding inequality is also trivially valid when p = ∞. Now set m =[

n+1p

]+1

and use (2.2.2) to obtain

sup|x|≥1

|x|m|∂ β f (x)| ≤ sup|x|≥1

Cn,m ∑|α |=m

|xα∂ β f (x)| ≤Cn,m ∑|α |≤m

ρα ,β ( f ) .

Conclusion (2.2.8) now follows immediately. This shows that convergence in S

implies convergence in Lp.

We now show that the Schwartz class is closed under certain operations.

Proposition 2.2.7. Let f , g be in S (Rn). Then f g and f ∗ g are in S (Rn). More-

over,

∂α( f ∗g) = (∂α f )∗g = f ∗ (∂αg) (2.2.9)

for all multi-indices α .

Proof. Fix f and g in S (Rn). Let e j be the unit vector (0, . . . ,1, . . . ,0) with 1 in the

jth entry and zeros in all the other entries. Since

f (y+he j)− f (y)

h− (∂ j f )(y)→ 0 (2.2.10)

as h→ 0, and since the expression in (2.2.10) is pointwise bounded by some constant

depending on f , the integral of the expression in (2.2.10) with respect to the measure

g(x− y)dy converges to zero as h → 0 by the Lebesgue dominated convergence

theorem. This proves (2.2.9) when α = (0, . . . ,1, . . . ,0). The general case follows by

repeating the previous argument and using induction.

We now show that the convolution of two functions in S is also in S . For each

N > 0 there is a constant CN such that

∣∣∣∣∫

Rnf (x− y)g(y)dy

∣∣∣∣≤CN

∫

Rn(1+ |x− y|)−N(1+ |y|)−N−n−1dy . (2.2.11)


Inserting the simple estimate

(1+ |x− y|)−N ≤ (1+ |y|)N(1+ |x|)−N

in (2.2.11) we obtain that

|( f ∗g)(x)| ≤CN(1+ |x|)−N

∫

Rn(1+ |y|)−n−1dy =C′N (1+ |x|)−N .

This shows that f ∗g decays like (1+ |x|)−N at infinity, but since N > 0 is arbitrary

it follows that f ∗g decays faster than the reciprocal of any polynomial.

Since ∂α( f ∗ g) = (∂α f ) ∗ g, replacing f by ∂α f in the previous argument, we

also conclude that all the derivatives of f ∗g decay faster than the reciprocal of any

polynomial at infinity. Using (2.2.7), we conclude that f ∗g is in S . Finally, the fact

that f g is in S follows directly from Leibniz’s rule (2.2.5) and (2.2.7).

2.2.2 The Fourier Transform of a Schwartz Function

The Fourier transform is often introduced as an operation on L1. In that setting,

problems of convergence arise when certain manipulations of functions are per-

formed. Also, Fourier inversion requires the additional assumption that the Fourier

transform is in L1. Here we initially introduce the Fourier transform on the space

of Schwartz functions. The rapid decay of Schwartz functions at infinity allows us

to develop its fundamental properties without encountering any convergence prob-

lems. The Fourier transform is a homeomorphism of the Schwartz class and Fourier

inversion holds in it. For these reasons, this class is a natural environment for it.

For x = (x1, . . . ,xn), y = (y1, . . . ,yn) in Rn we use the notation

x · y =n

∑j=1

x jy j .

Definition 2.2.8. Given f in S (Rn) we define

f (ξ ) =∫

Rnf (x)e−2πix·ξ dx .

We call f the Fourier transform of f .

Example 2.2.9. If f (x) = e−π|x|2

defined on Rn, then f (ξ ) = f (ξ ). To prove this,

observe that the function

s →∫ +∞

−∞e−π(t+is)2

dt, s ∈ R ,


defined on the line is constant (and thus equal to∫ +∞−∞ e−πt2

dt), since its derivative is

∫ +∞

−∞−2πi(t + is)e−π(t+is)2

dt =∫ +∞

−∞i

d

dt(e−π(t+is)2

)dt = 0 .

Using this fact, we calculate the Fourier transform of the function t → e−πt2on R

by completing the squares as follows:

∫

Re−πt2

e−2πitτdt =∫

Re−π(t+iτ)2

eπ(iτ)2

dt =

(∫ +∞

−∞e−πt2

dt

)e−πτ

2

= e−πτ2

,

where τ ∈ R, and we used that

∫ +∞

−∞e−t2

dt =√π , (2.2.12)

a fact that can be found in Appendix A.1.

Remark 2.2.10. It follows from the definition of the Fourier transform that if f is in

S (Rn) and g is in S (Rm), then

[ f (x1, . . . ,xn)g(xn+1, . . . ,xn+m)]= f (ξ1, . . . ,ξn)g(ξn+1, . . . ,ξn+m),

where the first denotes the Fourier transform on Rn+m. In other words, the Fourier

transform preserves separation of variables. Combining this observation with the

result in Example 2.2.9, we conclude that the function f (x) = e−π|x|2

defined on Rn

is equal to its Fourier transform.

We now continue with some properties of the Fourier transform. Before we do

this we introduce some notation. For a measurable function f on Rn, x ∈ Rn, and

a > 0 we define the translation, dilation, and reflection of f by

(τy f )(x) = f (x− y)

(δ a f )(x) = f (ax)

f (x) = f (−x).

(2.2.13)

Also recall the notation fa = a−nδ 1/a( f ) introduced in Definition 2.1.9.

Proposition 2.2.11. Given f , g in S (Rn), y∈Rn, b∈C, α a multi-index, and t > 0,

we have

(1)∥∥ f

∥∥L∞≤

∥∥ f∥∥

L1 ,

(2) f +g = f + g,

(3) b f = b f ,

(4)f =

˜f ,

(5) f =˜f ,

(6) τy f (ξ ) = e−2πiy·ξ f (ξ ),

(7) (e2πix·y f (x))(ξ ) = τy( f )(ξ ),

(8) (δ t f )= t−nδ t−1f = ( f )t ,

(9) (∂α f )(ξ ) = (2πiξ )α f (ξ ),

(10) (∂α f )(ξ ) = ((−2πix)α f (x))(ξ ),(11) f ∈S ,

(12) f ∗g = f g,

(13) f A(ξ )= f (Aξ ), where A is an orthogonal matrix and ξ is a column vector.

Proof. Property (1) follows directly from Definition 2.2.8. Properties (2)–(5) are

trivial. Properties (6)–(8) require a suitable change of variables but they are omitted.

Property (9) is proved by integration by parts (which is justified by the rapid decay

of the integrands):

(∂α f )(ξ ) =

∫

Rn(∂α f )(x)e−2πix·ξ dx

= (−1)|α |∫

Rnf (x)(−2πiξ )αe−2πix·ξ dx

= (2πiξ )α f (ξ ) .

To prove (10), let α = e j = (0, . . . ,1, . . . ,0), where all entries are zero except for

the jth entry, which is 1. Since

e−2πix·(ξ+he j)− e−2πix·ξ

h− (−2πix j)e

−2πix·ξ → 0 (2.2.14)

as h→ 0 and the preceding function is bounded by C|x| for all h and ξ , the Lebesgue

dominated convergence theorem implies that the integral of the function in (2.2.14)

with respect to the measure f (x)dx converges to zero. This proves (10) for α = e j.

For other α’s use induction. To prove (11) we use (9), (10), and (1) in the following

way:

∥∥xα(∂ β f )(x)∥∥

L∞=

(2π)|β |

(2π)|α |∥∥(∂α(xβ f (x)))

∥∥L∞≤ (2π)|β |

(2π)|α |∥∥∂α(xβ f (x))

∥∥L1 < ∞ .


Identity (12) follows from the following calculation:

f ∗g(ξ ) =∫

Rn

∫

Rnf (x− y)g(y)e−2πix·ξ dydx

=∫

Rn

∫

Rnf (x− y)g(y)e−2πi(x−y)·ξ e−2πiy·ξ dydx

=∫

Rng(y)

∫

Rnf (x− y)e−2πi(x−y)·ξdx e−2πiy·ξ dy

= f (ξ )g(ξ ),

where the application of Fubini’s theorem is justified by the absolute convergence

of the integrals. Finally, we prove (13). We have

f A(ξ ) =∫

Rnf (Ax)e−2πix·ξ dx

=∫

Rnf (y)e−2πiA−1y·ξ dy

=∫

Rnf (y)e−2πiAt y·ξ dy

=

∫

Rnf (y)e−2πiy·Aξ dy

= f (Aξ ) ,

where we used the change of variables y = Ax and the fact that |detA|= 1.

Corollary 2.2.12. The Fourier transform of a radial function is radial. Products and

convolutions of radial functions are radial.

Proof. Let ξ1, ξ2 in Rn with |ξ1|= |ξ2|. Then for some orthogonal matrix A we have

Aξ1 = ξ2. Since f is radial, we have f = f A. Then

f (ξ2) = f (Aξ1) = f A(ξ1) = f (ξ1),

where we used (13) in Proposition 2.2.11 to justify the second equality. Products and

convolutions of radial functions are easily seen to be radial.

2.2.3 The Inverse Fourier Transform and Fourier Inversion

We now define the inverse Fourier transform.

Definition 2.2.13. Given a Schwartz function f , we define

f∨(x) = f (−x),


for all x ∈ Rn. The operation

f → f∨

is called the inverse Fourier transform.

It is straightforward that the inverse Fourier transform shares the same properties

as the Fourier transform. One may want to list (and prove) properties for the inverse

Fourier transform analogous to those in Proposition 2.2.11.

We now investigate the relation between the Fourier transform and the inverse

Fourier transform. In the next theorem, we prove that one is the inverse operation of

the other. This property is referred to as Fourier inversion.

Theorem 2.2.14. Given f , g, and h in S (Rn), we have

(1)

∫

Rnf (x)g(x)dx =

∫

Rnf (x)g(x)dx ,

(2) (Fourier Inversion) ( f )∨ = f = ( f∨)∧ ,

(3) (Parseval’s relation)

∫

Rnf (x)h(x)dx =

∫

Rnf (ξ )h(ξ )dξ ,

(4) (Plancherel’s identity)∥∥ f

∥∥L2 =

∥∥ f∥∥

L2 =∥∥ f∨

∥∥L2 ,

(5)

∫

Rnf (x)h(x)dx =

∫

Rnf (x)h∨(x)dx .

Proof. (1) follows immediately from the definition of the Fourier transform and

Fubini’s theorem. To prove (2) we use (1) with

g(ξ ) = e2πiξ ·te−π|εξ |2

.

By Proposition 2.2.11 (7) and (8) and Example 2.2.9, we have that

g(x) =1

εne−π|(x−t)/ε |2 ,

which is an approximate identity. Now (1) gives

∫

Rnf (x)ε−ne−πε

−2|x−t|2 dx =

∫

Rnf (ξ )e2πiξ ·te−π|εξ |

2

dξ . (2.2.15)

Now let ε→ 0 in (2.2.15). The left-hand side of (2.2.15) converges to f (t) uniformly

on compact sets by Theorem 1.2.19. The right-hand side of (2.2.15) converges to

( f )∨(t) as ε → 0 by the Lebesgue dominated convergence theorem. We conclude

that ( f )∨= f on Rn. Replacing f by f and using the result just proved, we conclude

that ( f∨)∧ = f .

Note that if g = h, then Proposition 2.2.11 (5) and identity (2) imply that g = h.

Then (3) follows from (1) by expressing h in terms of g. Identity (4) is a trivial


consequence of (3). (Sometimes the polarized identity (3) is also referred to as

Plancherel’s identity.) Finally, (5) easily follows from (1) and (2) with g = h.

Next we have the following simple corollary of Theorem 2.2.14.

Corollary 2.2.15. The Fourier transform is a homeomorphism from S (Rn) onto

itself.

Proof. The continuity of the Fourier transform (and its inverse) follows from Exer-

cise 2.2.2, while Fourier inversion yields that this map is bijective.

2.2.4 The Fourier Transform on L1 +L2

We have defined the Fourier transform on S (Rn). We now extend this definition to

the space L1(Rn)+L2(Rn).We begin by observing that the Fourier transform given in Definition 2.2.8,

f (ξ ) =∫

Rnf (x)e−2πix·ξ dx ,

makes sense as a convergent integral for functions f ∈ L1(Rn). This allows us to

extend the definition of the Fourier transform on L1. Moreover, this operator satisfies

properties (1)–(8) as well as (12) and (13) in Proposition 2.2.11, with f ,g integrable.

We also define the inverse Fourier transform on L1 by setting f∨(x) = f (−x) for

f ∈ L1(Rn) and we note that analogous properties hold for it. One problem in this

generality is that when f is integrable, one may not necessarily have ( f )∨ = f a.e.

This inversion is possible when f is also integrable; see Exercise 2.2.6.

The integral defining the Fourier transform does not converge absolutely for func-

tions in L2(Rn); however, the Fourier transform has a natural definition in this space

accompanied by an elegant theory. In view of the result in Exercise 2.2.8, the Fourier

transform is an L2 isometry on L1∩L2, which is a dense subspace of L2. By density,

there is a unique bounded extension of the Fourier transform on L2. Let us denote

this extension by F . Then F is also an isometry on L2, i.e.,

∥∥F ( f )∥∥

L2 =∥∥ f

∥∥L2

for all f ∈ L2(Rn), and any sequence of functions fN ∈ L1(Rn)∩L2(Rn) converging

to a given f in L2(Rn) satisfies

∥∥ fN−F ( f )∥∥

L2 → 0 , (2.2.16)

as N → ∞. In particular, the sequence of functions fN(x) = f (x)χ|x|≤N yields that

fN(ξ ) =∫

|x|≤Nf (x)e−2πix·ξ dx (2.2.17)

converges to F ( f )(ξ ) in L2 as N → ∞. If f is both integrable and square inte-

grable, the expressions in (2.2.17) also converge to f (ξ ) pointwise. Also, in view of

Theorem 1.1.11 and (2.2.16), there is a subsequence of fN that converges to F ( f )

pointwise a.e. Consequently, for f in L1(Rn)∩L2(Rn) the expressions f and F ( f )

coincide pointwise a.e. For this reason we often adopt the notation f to denote the

Fourier transform of functions f in L2 as well.

In a similar fashion, we let F ′ be the isometry on L2(Rn) that extends the op-

erator f → f∨, which is an L2 isometry on L1 ∩ L2; the last statement follows

by adapting the result of Exercise 2.2.8 to the inverse Fourier transform. Since

ϕ∨(x) = ϕ(−x) for ϕ in the Schwartz class, which is dense in L2 (Exercise 2.2.5),

it follows that F ′( f )(x) = F ( f )(−x) for all f ∈ L2 and almost all x ∈ Rn. The

operators F and F ′ are L2-isometries that satisfy F ′ F = F F ′ = Id on the

Schwartz space. By density this identity also holds for L2 functions and implies that

F and F ′ are injective and surjective mappings from L2 to itself; consequently, F ′

coincides with the inverse operator F−1 of F : L2 → L2, and Fourier inversion

f = F−1 F ( f ) = F F

−1( f ) a.e.

holds on L2.

Having set down the basic facts concerning the action of the Fourier transform

on L1 and L2, we extend its definition on Lp for 1 1 and f2 = f χ| f |≤1. The

definition of f is independent of the choice of f1 and f2, for if f1 + f2 = h1 +h2 for

f1,h1 ∈ L1(Rn) and f2,h2 ∈ L2(Rn), we have f1−h1 = h2− f2 ∈ L1(Rn)∩L2(Rn).Since these functions are equal on L1(Rn)∩ L2(Rn), their Fourier transforms are

also equal, and we obtain f1− h1 = h2− f2, which yields f1 + f2 = h1 +h2. We

have the following result concerning the action of the Fourier transform on Lp.

Proposition 2.2.16. (Hausdorff–Young inequality) For every function f in Lp(Rn)we have the estimate ∥∥ f

∥∥Lp′ ≤

∥∥ f∥∥

Lp

whenever 1≤ p≤ 2.

Proof. This follows easily from Theorem 1.3.4. Interpolate between the estimates

‖ f ‖L∞ ≤ ‖ f‖L1 (Proposition 2.2.11 (1)) and ‖ f ‖L2 ≤ ‖ f‖L2 to obtain ‖ f ‖Lp′ ≤

‖ f‖Lp . We conclude that the Fourier transform is a bounded operator from Lp(Rn)

to Lp′(Rn) with norm at most 1 when 1≤ p≤ 2.

Next, we are concerned with the behavior of the Fourier transform at infinity.

Proposition 2.2.17. (Riemann–Lebesgue lemma) For a function f in L1(Rn) we

have that

| f (ξ )| → 0 as |ξ | → ∞ .

Proof. Consider the function χ[a,b] on R. A simple computation gives

χ[a,b](ξ ) =∫ b

ae−2πixξ dx =

e−2πiξa− e−2πiξb

2πiξ,

which tends to zero as |ξ | → ∞. Likewise, if g =∏nj=1 χ[a j ,b j ] on Rn, then

g(ξ ) =n

∏j=1

e−2πiξ ja j − e−2πiξ jb j

2πiξ j

.

Given a ξ = (ξ1, . . . ,ξn) = 0, there is j0 such that |ξ j0 |> |ξ |/√

n. Then

∣∣∣∣n

∏j=1

e−2πiξ ja j − e−2πiξ jb j

2πiξ j

∣∣∣∣≤2√

n

2π|ξ | sup1≤ j0≤n

∏j = j0

(b j−a j)

which also tends to zero as |ξ | → ∞ in Rn.

Given a general integrable function f on Rn and ε > 0, there is a simple function

h, which is a finite linear combination of characteristic functions of rectangles (like

g), such that ‖ f − h‖L1 < ε2. Then there is an M is such that for |ξ | > M we have

|h(ξ )|< ε2. It follows that

| f (ξ )| ≤ | f (ξ )− h(ξ )|+ |h(ξ )| ≤∥∥ f −h

∥∥L1 + |h(ξ )|<

ε

2+ε

2,

provided |ξ |> M. This implies that | f (ξ )| → 0 as |ξ | → ∞.

A different proof can be given by taking the function h in the preceding paragraph

to be a Schwartz function and using that Schwartz functions are dense in L1(Rn);see Exercise 2.2.5 about the last assertion.

We end this section with an example that illustrates some of the practical uses of

the Fourier transform.

Example 2.2.18. We would like to find a Schwartz function f (x1,x2,x3) on R3 that

satisfies the partial differential equation

f (x)+∂ 21 ∂

22 ∂

43 f (x)+4i∂ 2

1 f (x)+∂ 72 f (x) = e−π|x|

2

.

Taking the Fourier transform on both sides of this identity and using Proposition

2.2.11 (2), (9) and the result of Example 2.2.9, we obtain

f (ξ )[1+(2πiξ1)

2(2πiξ2)2(2πiξ3)

4 +4i(2πiξ1)2 +(2πiξ2)

7]= e−π|ξ |

2

.

Let p(ξ ) = p(ξ1,ξ2,ξ3) be the polynomial inside the square brackets. We observe

that p(ξ ) has no real zeros and we may therefore write

f (ξ ) = e−π|ξ |2

p(ξ )−1 =⇒ f (x) =(e−π|ξ |

2

p(ξ )−1)∨

(x) .

In general, let

P(ξ ) = ∑|α |≤N

Cαξα

be a polynomial in Rn with constant complex coefficients Cα indexed by multi-

indices α . If P(2πiξ ) has no real zeros, and u is in S (Rn), then the partial differ-

ential equation

P(∂ ) f = ∑|α |≤N

Cα∂α f = u

is solved as before to give

f =(u(ξ )P(2πiξ )−1

)∨.

Since P(2πiξ ) has no real zeros and u ∈S (Rn), the function

u(ξ )P(2πiξ )−1

is smooth and therefore a Schwartz function. Then f is also in S (Rn) by Proposition

2.2.11 (11).

Exercises

2.2.1. (a) Construct a Schwartz function supported in the unit ball of Rn.

(b) Construct a C ∞0 (Rn) function equal to 1 on the annulus 1≤ |x| ≤ 2 and vanishing

off the annulus 1/2≤ |x| ≤ 4.

(c) Construct a nonnegative nonzero Schwartz function f on Rn whose Fourier

transform is nonnegative and compactly supported.[Hint: Part (a): Try the construction in dimension one first using the C ∞ function

η(x) = e−1/x for x > 0 and η(x) = 0 for x < 0. Part (c): Take f = |φ ∗ φ |2, where φ

is odd, real-valued, and compactly supported; here φ(x) = φ(−x).]

2.2.2. If fk, f ∈S (Rn) and fk → f in S (Rn), then fk → f and f∨k → f∨ in S (Rn).

2.2.3. Find the spectrum (i.e., the set of all eigenvalues of the Fourier transform),

that is, all complex numbers λ for which there exist nonzero functions f such that

f = λ f .

[Hint: Apply the Fourier transform three times to the preceding identity. Consider

the functions xe−πx2, (a+ bx2)e−πx2

, and (cx+ dx3)e−πx2for suitable a,b,c,d to

show that all fourth roots of unity are indeed eigenvalues of the Fourier transform.]

2.2.4. Use the idea of the proof of Proposition 2.2.7 to show that if the functions f ,

g defined on Rn satisfy | f (x)| ≤ A(1+ |x|)−M and |g(x)| ≤ B(1+ |x|)−N for some

M,N > n, then

|( f ∗g)(x)| ≤ ABC(1+ |x|)−L ,

where L = min(N,M) and C =C(N,M)> 0.

2.2.5. Show that C ∞0 (Rn) is dense on Lp(Rn) for 0 < p < ∞ but not for p = ∞.[

Hint: Use a smooth approximate identity when p ≥ 1. Reduce the case p < 1 to

p = 1.]

2.2.6. (a) Prove that if f ∈ L1, then f is uniformly continuous on Rn.

(b) Prove that for f ,g ∈ L1(Rn) we have

∫

Rnf (x)g(x)dx =

∫

Rnf (y)g(y)dy .

(c) Take g(x) = ε−ne−πε−2|x−t|2 in (b) and let ε→ 0 to prove that if f and f are both

in L1, then ( f )∨ = f a.e. This fact is called Fourier inversion on L1.

2.2.7. (a) Prove that if a function f in L1(Rn)∩L∞(Rn) is continuous at 0, then

limε→0

∫

Rnf (x)e−π|εx|2dx = f (0) .

(b) Let f ∈ L1(Rn)∩L∞(Rn) be continuous at zero and satisfy f ≥ 0. Show that f

is in L1 and conclude that Fourier inversion holds at zero f (0) = ‖ f ‖L1 , and also

f = ( f )∨ a.e. in general.[Hint: Part (a): Let g(x) = e−π|εx|2 in Exercise 2.2.6(b) and use Theorem 1.2.19 (2).

]

2.2.8. Given f in L1(Rn)∩L2(Rn), without appealing to density, prove that

∥∥ f∥∥

L2 =∥∥ f

∥∥L2 .

[Hint: Let h = f ∗ f , where f (x) = f (−x) and the bar indicates complex conjuga-

tion. Then h ∈ L1(Rn)∩L∞(Rn), h = | f |2 ≥ 0, and h is continuous at zero. Exercise

2.2.7(b) yields ‖ f ‖2L2 = ‖h‖L1 = h(0) =

∫

Rnf (x) f (−x)dx = ‖ f‖2

L2 .]

2.2.9. (a) Prove that for all 0 < ε < t < ∞ we have

∣∣∣∣∫ t

ε

sin(ξ )

ξdξ

∣∣∣∣≤ 4 .

(b) If f is an odd L1 function on the line, conclude that for all t > ε > 0 we have

∣∣∣∣∫ t

ε

f (ξ )

ξdξ

∣∣∣∣≤ 4∥∥ f

∥∥L1 .

(c) Let g(ξ ) be a continuous odd function that is equal to 1/ log(ξ ) for ξ ≥ 2. Show

that there does not exist an L1 function whose Fourier transform is g.


2.2.10. Let f be in L1(R). Prove that

∫ +∞

−∞f(

x− 1

x

)dx =

∫ +∞

−∞f (u)du .

[Hint: For x∈ (−∞,0) use the change of variables u = x− 1

xor x = 1

2

(u−√

4+u2).

For x ∈ (0,∞) use the change of variables u = x− 1x

or x = 12

(u+

√4+u2

).]

2.2.11. (a) Use Exercise 2.2.10 with f (x) = e−tx2to obtain the subordination

identity

e−2t =1√π

∫ ∞

0e−y−t2/y dy√

y, where t > 0.

(b) Set t = π|x| and integrate with respect to e−2πiξ ·xdx to prove that

(e−2π|x|)(ξ ) = Γ ( n+12)

πn+1

2

1

(1+ |ξ |2) n+12

.

This calculation gives the Fourier transform of the Poisson kernel.

2.2.12. Let 1≤ p≤ ∞ and let p′ be its dual index.

(a) Prove that Schwartz functions f on the line satisfy the estimate

∥∥ f∥∥2

L∞≤ 2

∥∥ f∥∥

Lp

∥∥ f ′∥∥

Lp′ .

(b) Prove that all Schwartz functions f on Rn satisfy the estimate

∥∥ f∥∥2

L∞≤ ∑|α+β |=n

∥∥∂α f∥∥

Lp

∥∥∂ β f∥∥

Lp′ ,

where the sum is taken over all pairs of multi-indices α and β whose sum has size n.[Hint: Part (a): Write f (x)2 =

∫ x−∞

ddt

f (t)2 dt.]

2.2.13. The uncertainty principle says that the position and the momentum of a

particle cannot be simultaneously localized. Prove the following inequality, which

presents a quantitative version of this principle:

∥∥ f∥∥2

L2(Rn)≤ 4π

ninf

y∈Rn

[∫

Rn|x− y|2| f (x)|2 dx

]12

infz∈Rn

[∫

Rn|ξ − z|2| f (ξ )|2 dξ

]12

,

where f is a Schwartz function on Rn (or an L2 function with sufficient decay at

infinity).[Hint: Let y be in Rn. Start with

∥∥ f∥∥2

L2 =1

n

∫

Rnf (x) f (x)

n

∑j=1

∂

∂x j

(x j− y j)dx ,

2.3 The Class of Tempered Distributions 119

integrate by parts, apply the Cauchy–Schwarz inequality, Plancherel’s identity, and

the identity ∑nj=1 |∂ j f (ξ )|2 = 4π2|ξ |2| f (ξ )|2 for all ξ ∈ Rn. Then replace f (x) by

f (x)e2πix·z.]

2.2.14. Let −∞< α < n2< β <+∞. Prove the validity of the following inequality:

∥∥g∥∥

L1(Rn)≤C

∥∥|x|αg(x)∥∥β−n/2

β−αL2(Rn)

∥∥|x|βg(x)∥∥

n/2−αβ−α

L2(Rn)

for some constant C =C(n,α,β ) independent of g.[Hint: First prove ‖g‖L1 ≤C‖|x|αg(x)‖L2 + ‖|x|βg(x)‖L2 and then replace g(x) by

g(λx) for some suitable λ > 0.]

2.3 The Class of Tempered Distributions

The fundamental idea of the theory of distributions is that it is generally easier to

work with linear functionals acting on spaces of “nice” functions than to work with

“bad” functions directly. The set of “nice” functions we consider is closed under

the basic operations in analysis, and these operations are extended to distributions

by duality. This wonderful interpretation has proved to be an indispensable tool that

has clarified many situations in analysis.

2.3.1 Spaces of Test Functions

We recall the space C ∞0 (Rn) of all smooth functions with compact support, and

C ∞(Rn) of all smooth functions on Rn. We are mainly interested in the three spaces

of “nice” functions on Rn that are nested as follows:

C∞0 (Rn)⊆S (Rn)⊆ C

∞(Rn) .

Here S (Rn) is the space of Schwartz functions introduced in Section 2.2.

Definition 2.3.1. We define convergence of sequences in these spaces. We say that

fk → f in C∞ ⇐⇒ fk, f ∈ C

∞ and limk→∞

sup|x|≤N

|∂α( fk− f )(x)|= 0

∀ α multi-indices and all N = 1,2, . . . .

fk → f in S ⇐⇒ fk, f ∈S and limk→∞

supx∈Rn

|xα∂ β ( fk− f )(x)|= 0

∀ α,β multi-indices.

fk → f in C∞0 ⇐⇒ fk, f ∈ C

∞0 , support( fk)⊆ B for all k, B compact,

and limk→∞

∥∥∂α( fk− f )∥∥

L∞= 0 ∀ α multi-indices.


It follows that convergence in C ∞0 (Rn) implies convergence in S (Rn), which in

turn implies convergence in C ∞(Rn).

Example 2.3.2. Let ϕ be a nonzero C ∞0 function on R. We call such functions

smooth bumps. Define the sequence of smooth bumps ϕk(x) = ϕ(x− k)/k. Then

ϕk(x) does not converge to zero in C ∞0 (R), even though ϕk (and all of its deriva-

tives) converge to zero uniformly. Furthermore, we see that ϕk does not converge to

any function in S (R). Clearly ϕk → 0 in C ∞(R).

The space C ∞(Rn) is equipped with the family of seminorms

ρα ,N( f ) = sup|x|≤N

|(∂α f )(x)|, (2.3.1)

where α ranges over all multi-indices and N ranges over Z+. It can be shown that

C ∞(Rn) is complete with respect to this countable family of seminorms, i.e., it is a

Frechet space. However, it is true that C ∞0 (Rn) is not complete with respect to the

topology generated by this family of seminorms.

The topology of C ∞0 given in Definition 2.3.1 is the inductive limit topology, and

under this topology it is complete. Indeed, letting C ∞0 (B(0,k)) be the space of all

smooth functions with support in B(0,k), then C ∞0 (Rn) is equal to

⋃∞k=1 C ∞

0 (B(0,k))and each space C ∞

0 (B(0,k)) is complete with respect to the topology generated by

the family of seminorms ρα ,N ; hence so is C ∞0 (Rn). Nevertheless, C ∞

0 (Rn) is not

metrizable. Details on the topologies of these spaces can be found in [286].

2.3.2 Spaces of Functionals on Test Functions

The dual spaces (i.e., the spaces of continuous linear functionals on the sets of test

functions) we introduced is denoted by

(C ∞0 (Rn))′ = D

′(Rn) ,

(S (Rn))′ = S′(Rn) ,

(C ∞(Rn))′ = E′(Rn) .

By definition of the topologies on the dual spaces, we have

Tk → T in D′ ⇐⇒ Tk,T ∈D

′ and Tk( f )→ T ( f ) for all f ∈ C∞0 .

Tk → T in S′ ⇐⇒ Tk,T ∈S

′ and Tk( f )→ T ( f ) for all f ∈S .

Tk → T in E′ ⇐⇒ Tk,T ∈ E

′ and Tk( f )→ T ( f ) for all f ∈ C∞.

The dual spaces are nested as follows:

E′(Rn)⊆S

′(Rn)⊆D′(Rn) .

Definition 2.3.3. Elements of the space D ′(Rn) are called distributions. Elements of

S ′(Rn) are called tempered distributions. Elements of the space E ′(Rn) are called

distributions with compact support.

Before we discuss some examples, we give alternative characterizations of distri-

butions, which are very useful from the practical point of view. The action of a

distribution u on a test function f is represented in either one of the following two

ways: ⟨u, f

⟩= u( f ) .

Proposition 2.3.4. (a) A linear functional u on C ∞0 (Rn) is a distribution if and only

if for every compact K ⊆ Rn, there exist C > 0 and an integer m such that

∣∣⟨u, f⟩∣∣≤C ∑

|α |≤m

∥∥∂α f∥∥

L∞, for all f ∈ C

∞ with support in K. (2.3.2)

(b) A linear functional u on S (Rn) is a tempered distribution if and only if there

exist C > 0 and k, m integers such that

∣∣⟨u, f⟩∣∣≤C ∑

|α |≤m|β |≤k

ρα ,β ( f ), for all f ∈S (Rn). (2.3.3)

(c) A linear functional u on C ∞(Rn) is a distribution with compact support if and

only if there exist C > 0 and N, m integers such that

∣∣⟨u, f⟩∣∣≤C ∑

|α |≤m

ρα ,N( f ), for all f ∈ C∞(Rn). (2.3.4)

The seminorms ρα ,β and ρα ,N are defined in (2.2.6) and (2.3.1), respectively.

Proof. We prove only (2.3.3), since the proofs of (2.3.2) and (2.3.4) are similar. It is

clear that (2.3.3) implies continuity of u. Conversely, it was pointed out in Section

2.2 that the family of sets f ∈ S (Rn) : ρα ,β ( f ) < δ, where α , β are multi-

indices and δ > 0, forms a subbasis for the topology of S . Thus if u is a continuous

functional on S , there exist integers k, m and a δ > 0 such that

|α| ≤ m, |β | ≤ k, and ρα ,β ( f )< δ =⇒∣∣⟨u, f

⟩∣∣≤ 1. (2.3.5)

We see that (2.3.3) follows from (2.3.5) with C = 1/δ .

Examples 2.3.5. We now discuss some important examples.

1. The Dirac mass at the origin δ0. This is defined for ϕ ∈ C ∞(Rn) by

⟨δ0,ϕ

⟩= ϕ(0).


We claim that δ0 is in E ′. To see this we observe that if ϕk → ϕ in C ∞ then⟨δ0,ϕk

⟩→

⟨δ0,ϕ

⟩. The Dirac mass at a point a ∈ Rn is defined similarly by

⟨δa,ϕ

⟩= ϕ(a).

2. Some functions g can be thought of as distributions via the identification g → Lg,

where Lg is the functional

Lg(ϕ) =∫

Rnϕ(x)g(x)dx .

Here are some examples: The function 1 is in S ′ but not in E ′. Compactly sup-

ported integrable functions are in E ′. The function e|x|2

is in D ′ but not in S ′.3. Functions in L1

loc are distributions. To see this, first observe that if g ∈ L1loc, then

the integral

Lg(ϕ) =∫

Rnϕ(x)g(x)dx

is well defined for all ϕ ∈ D and satisfies |Lg(ϕ)| ≤(∫

K |g(x)|dx)‖ϕ‖L∞ for all

smooth functions ϕ supported in the compact set K.

4. Functions in Lp, 1≤ p≤ ∞, are tempered distributions, but may not in E ′ unless

they have compact support.

5. Any finite Borel measure µ is a tempered distribution via the identification

Lµ(ϕ) =∫

Rnϕ(x)dµ(x) .

To see this, observe that ϕk → ϕ in S implies that Lµ(ϕk)→ Lµ(ϕ). Finite Borel

measures may not be distributions with compact support.

6. Every function g that satisfies |g(x)| ≤ C(1+ |x|)k, for some real number k, is a

tempered distribution. To see this, observe that

∣∣Lg(ϕ)∣∣≤ sup

x∈Rn(1+ |x|)m|ϕ(x)|

∫

Rn(1+ |x|)k−mdx ,

where m> n+k and the expression supx∈Rn(1+ |x|)m|ϕ(x)| is bounded by a finite

sum of Schwartz seminorms ρα ,β (ϕ).7. The function log |x| is a tempered distribution; indeed for any ϕ ∈ S (Rn), the

integral of ϕ(x) log |x| is bounded by a finite number of Schwartz seminorms of

ϕ . More generally, any function that is integrable on a ball |x| ≤M and for some

C > 0 satisfies |g(x)| ≤C(1+ |x|)k for |x| ≥M, is a tempered distribution.

8. Here is an example of a compactly supported distribution on R that is neither a

locally integrable function nor a finite Borel measure:

⟨u,ϕ

⟩= lim

ε→0

∫

ε≤|x|≤1

ϕ(x)dx

x= lim

ε→0

∫

ε≤|x|≤1

(ϕ(x)−ϕ(0))dx

x.

We have that |〈u,ϕ〉| ≤ 2‖ϕ ′‖L∞([−1,1]) and notice that ‖ϕ ′‖L∞([−1,1]) is a ρα ,Nseminorm of ϕ .


2.3.3 The Space of Tempered Distributions

Having set down the basic definitions of distributions, we now focus our study on the

space of tempered distributions. These distributions are the most useful in harmonic

analysis. The main reason for this is that the subject is concerned with boundedness

of translation-invariant operators, and every such bounded operator from Lp(Rn) to

Lq(Rn) is given by convolution with a tempered distribution. This fact is shown in

Section 2.5.

Suppose that f and g are Schwartz functions and α a multi-index. Integrating by

parts |α| times, we obtain

∫

Rn(∂α f )(x)g(x)dx = (−1)|α |

∫

Rnf (x)(∂αg)(x)dx. (2.3.6)

If we wanted to define the derivative of a tempered distribution u, we would have to

give a definition that extends the definition of the derivative of the function and that

satisfies (2.3.6) for g in S ′ and f ∈S if the integrals in (2.3.6) are interpreted as

actions of distributions on functions. We simply use equation (2.3.6) to define the

derivative of a distribution.

Definition 2.3.6. Let u ∈S ′ and α a multi-index. Define

⟨∂αu, f

⟩= (−1)|α |

⟨u,∂α f

⟩. (2.3.7)

If u is a function, the derivatives of u in the sense of distributions are called distri-

butional derivatives.

In view of Theorem 2.2.14, it is natural to give the following:

Definition 2.3.7. Let u ∈ S ′. We define the Fourier transform u and the inverse

Fourier transform u∨ of a tempered distribution u by

⟨u, f

⟩=

⟨u, f

⟩and

⟨u∨, f

⟩=

⟨u, f∨

⟩, (2.3.8)

for all f in S .

Example 2.3.8. We observe that δ0 = 1. More generally, for any multi-index α we

have

(∂αδ0)∧ = (2πix)α .

To see this, observe that for all f ∈S we have

⟨(∂αδ0)

∧ , f⟩=

⟨∂αδ0 , f

⟩

= (−1)|α |⟨δ0 , ∂

α f⟩

= (−1)|α |⟨δ0 , ((−2πix)α f (x))∧

⟩

= (−1)|α |((−2πix)α f (x))∧ (0)


= (−1)|α |∫

Rn(−2πix)α f (x)dx

=∫

Rn(2πix)α f (x)dx .

This calculation indicates that (∂αδ0)∧ can be identified with the function (2πix)α .

Example 2.3.9. Recall that for x0 ∈ Rn, δx0( f ) =

⟨δx0

, f⟩= f (x0). Then

⟨δx0

,h⟩=

⟨δx0

, h⟩= h(x0) =

∫

Rnh(x)e−2πix·x0 dx, h ∈S (Rn) ,

that is, δx0can be identified with the function x → e−2πix·x0 . In particular, δ0 = 1.

Example 2.3.10. The function e|x|2

is not in S ′(Rn) and therefore its Fourier trans-

form is not defined as a distribution. However, the Fourier transform of any locally

integrable function with polynomial growth at infinity is defined as a tempered dis-

tribution.

Now observe that the following are true whenever f , g are in S .

∫

Rng(x) f (x− t)dx =

∫

Rng(x+ t) f (x)dx ,

∫

Rng(ax) f (x)dx =

∫

Rng(x)a−n f (a−1x)dx ,

∫

Rng(x) f (x)dx =

∫

Rng(x) f (x)dx ,

(2.3.9)

for all t ∈Rn and a > 0. Recall now the definitions of τ t , δ a, and ˜ given in (2.2.13).

Motivated by (2.3.9), we give the following:

Definition 2.3.11. The translation τ tu, the dilation δ au, and the reflection u of a

tempered distribution u are defined as follows:

⟨τ tu, f

⟩=

⟨u,τ−t f

⟩, (2.3.10)

⟨δ au, f

⟩=

⟨u,a−nδ 1/a f

⟩, (2.3.11)

⟨u, f

⟩=

⟨u, f

⟩, (2.3.12)

for all t ∈ Rn and a > 0. Let A be an invertible matrix. The composition of a distri-

bution u with an invertible matrix A is the distribution

⟨uA,ϕ

⟩= |det A|−1

⟨u,ϕA−1⟩

, (2.3.13)

where ϕA−1(x) = ϕ(A−1x).


It is easy to see that the operations of translation, dilation, reflection, and differ-

entiation are continuous on tempered distributions.

Example 2.3.12. The Dirac mass at the origin δ0 is equal to its reflection, while

δ aδ0 = a−nδ0. Also, τxδ0 = δx for any x ∈ Rn.

Now observe that for f , g, and h in S we have

∫

Rn(h∗g)(x) f (x)dx =

∫

Rng(x)(h∗ f )(x)dx . (2.3.14)

Motivated by (2.3.14), we define the convolution of a function with a tempered dis-

tribution as follows:

Definition 2.3.13. Let u ∈S ′ and h ∈S . Define the convolution h∗u by

⟨h∗u, f

⟩=

⟨u, h∗ f

⟩, f ∈S . (2.3.15)

Example 2.3.14. Let u = δx0and f ∈S . Then f ∗δx0

is the function x → f (x−x0),for when h ∈S , we have

⟨f ∗δx0

,h⟩=

⟨δx0

, f ∗h⟩= ( f ∗h)(x0) =

∫

Rnf (x− x0)h(x)dx .

It follows that convolution with δ0 is the identity operator.

We now define the product of a function and a distribution.

Definition 2.3.15. Let u∈S ′ and let h be a C ∞ function that has at most polynomial

growth at infinity and the same is true for all of its derivatives. This means that for

all α it satisfies |(∂αh)(x)| ≤ Cα(1+ |x|)kα for some Cα ,kα > 0. Then define the

product hu of h and u by

⟨hu, f

⟩=

⟨u,h f

⟩, f ∈S . (2.3.16)

Note that h f is in S and thus (2.3.16) is well defined. The product of an arbitrary

C ∞ function with a tempered distribution is not defined.

We observe that if a function g is supported in a set K, then for all f ∈ C ∞0 (Kc)

we have ∫

Rnf (x)g(x)dx = 0 . (2.3.17)

Moreover, the support of g is the intersection of all closed sets K with the property

(2.3.17) for all f in C ∞0 (Kc). Motivated by the preceding observation we give the

following:

Definition 2.3.16. Let u be in D ′(Rn). The support of u (supp u) is the intersection

of all closed sets K with the property

ϕ ∈ C∞0 (Rn), suppϕ ⊆ Rn \K =⇒

⟨u,ϕ

⟩= 0 . (2.3.18)


Distributions with compact support are exactly those whose support (as defined

in the previous definition) is a compact set. To prove this assertion, we start with a

distribution u with compact support as defined in Definition 2.3.3. Then there exist

C,N,m > 0 such that (2.3.4) holds. For a C ∞ function f whose support is contained

in B(0,N)c, the expression on the right in (2.3.4) vanishes and we must therefore

have 〈u, f 〉= 0. This shows that the support of u is contained in B(0,N) hence it is

bounded, and since it is already closed (as an intersection of closed sets), it must be

compact. Conversely, if the support of u as defined in Definition 2.3.16 is a compact

set, then there exists an N > 0 such that supp u is contained in B(0,N). We take a

smooth function η that is equal to 1 on B(0,N) and vanishes off B(0,N +1). Then

for h∈C ∞0 the support of h(1−η) does not meet the support of u, and we must have

⟨u,h

⟩=

⟨u,hη

⟩+

⟨u,h(1−η)

⟩=

⟨u,hη

⟩.

The distribution u can be thought of as an element of E ′ by defining for f ∈C ∞(Rn)

⟨u, f

⟩=

⟨u, fη

⟩.

Taking m to be the integer that corresponds to the compact set K = B(0,N +1)in (2.3.2), and using that the L∞ norm of ∂α( fη) is controlled by a finite sum of

seminorms ρα ,N+1( f ) with |α| ≤ m, we obtain the validity of (2.3.4) for f ∈ C ∞.

Example 2.3.17. The support of the Dirac mass at x0 is the set x0.

Along the same lines, we give the following definition:

Definition 2.3.18. We say that a distribution u in D ′(Rn) coincides with the function

h on an open set Ω if

⟨u, f

⟩=

∫

Rnf (x)h(x)dx for all f in C

∞0 (Ω). (2.3.19)

When (2.3.19) occurs we often say that u agrees with h away from Ω c.

This definition implies that the support of the distribution u− h is contained in

the set Ω c.

Example 2.3.19. The distribution |x|2 +δa1+δa2

, where a1, a2 are in Rn, coincides

with the function |x|2 on any open set not containing the points a1 and a2. Also, the

distribution in Example 2.3.5 (8) coincides with the function x−1χ|x|≤1 away from

the origin in the real line.

Having ended the streak of definitions regarding operations with distributions,

we now discuss properties of convolutions and Fourier transforms.


Theorem 2.3.20. If u ∈S ′ and ϕ ∈S , then ϕ ∗u is a C ∞ function and

(ϕ ∗u)(x) = 〈u,τxϕ〉

for all x ∈ Rn. Moreover, for all multi-indices α there exist constants Cα ,kα > 0

such that

|∂α(ϕ ∗u)(x)| ≤Cα(1+ |x|)kα .

Furthermore, if u has compact support, then ϕ ∗u is a Schwartz function.

Proof. Let ψ be in S (Rn). We have

⟨ϕ ∗u,ψ

⟩=

⟨u, ϕ ∗ψ

⟩

= u

(∫

Rnϕ( · − y)ψ(y)dy

)

= u

(∫

Rn(τyϕ)( ·)ψ(y)dy

)(2.3.20)

=

∫

Rn

⟨u,τyϕ

⟩ψ(y)dy,

where the last step is justified by the continuity of u and by the fact that the Riemann

sums of the inner integral in (2.3.20) converge to that integral in the topology of S ,

a fact that will be justified later. This calculation identifies the function ϕ ∗u as

(ϕ ∗u)(x) =⟨u,τxϕ

⟩. (2.3.21)

We now show that (ϕ ∗ u)(x) is a C ∞ function. Let e j = (0, . . . ,1, . . . ,0) with 1

in the jth entry and zero elsewhere. Then

τ−he j(ϕ ∗u)(x)− (ϕ ∗u)(x)

h= u

(τ−he j(τxϕ)− τxϕ

h

)→

⟨u,τx(∂ jϕ)

⟩

by the continuity of u and the fact that(τ−he j(τxϕ)− τxϕ

)/h tends to ∂ jτ

xϕ =τx(∂ jϕ) in S as h→ 0; see Exercise 2.3.5 (a). The same calculation for higher-order

derivatives shows that ϕ ∗u∈C ∞ and that ∂ γ(ϕ ∗u) = (∂ γϕ)∗u for all multi-indices

γ . It follows from (2.3.3) that for some C, m, and k we have

|∂α(ϕ ∗u)(x)| ≤C ∑|γ |≤m|β |≤k

supy∈Rn

|yγτx(∂α+β ϕ)(y)|

=C ∑|γ |≤m|β |≤k

supy∈Rn

|(x+ y)γ(∂α+β ϕ)(y)|

≤Cm ∑|β |≤k

supy∈Rn

(1+ |x|m + |y|m)|(∂α+β ϕ)(y)| ,

(2.3.22)

and this clearly implies that ∂α(ϕ ∗u) grows at most polynomially at infinity.


We now indicate why ϕ ∗u is Schwartz whenever u has compact support. Apply-

ing estimate (2.3.4) to the function y → ϕ(x− y) yields that

∣∣⟨u,ϕ(x−·)⟩∣∣= |(ϕ ∗u)(x)| ≤C ∑

|α |≤m

sup|y|≤N

|∂αy ϕ(x− y)|

for some constants C,m,N. Since for |x| ≥ 2N we have

|∂αy ϕ(x− y)| ≤Cα ,M(1+ |x− y|)−M ≤Cα ,M,N(1+ |x|)−M,

it follows that ϕ ∗u decays rapidly at infinity. Since ∂ γ(ϕ ∗u) = (∂ γϕ)∗u, the same

argument yields that all the derivatives of ϕ ∗u decay rapidly at infinity; hence ϕ ∗u

is a Schwartz function. Incidentally, this argument actually shows that any Schwartz

seminorm of ϕ ∗u is controlled by a finite sum of Schwartz seminorms of ϕ .

We now return to the point left open concerning the convergence of the Riemann

sums in (2.3.20) in the topology of S (Rn). For each N = 1,2, . . . , consider a parti-

tion of [−N,N]n into (2N2)n cubes Qm of side length 1/N and let ym be the center

of each Qm. For multi-indices α,β , we must show that

DN(x) =(2N2)n

∑m=1

xα∂ βx ϕ(x− ym)ψ(ym)|Qm|−∫

Rnxα∂ βx ϕ(x− y)ψ(y)dy

converges to zero in L∞(Rn) as N → ∞. We have

xα∂ βx ϕ(x− ym)ψ(ym)|Qm|−∫

Qm

xα∂ βx ϕ(x− y)ψ(y)dy

=∫

Qm

xα(y− ym) ·∇(∂ βx ϕ(x−·)ψ

)(ξ )dy

for some ξ = y+θ(ym− y), where θ ∈ [0,1]. Distributing the gradient to both fac-

tors, we see that the last integrand is at most

C |x||α |√

n

N

1

(1+ |x−ξ |)M/2

1

(2+ |ξ |)M

for M large (pick M > 2|α|), which in turn is at most

C′ |x||α |√

n

N

1

(1+ |x|)M/2

1

(2+ |ξ |)M/2≤C′ |x||α |

√n

N

1

(1+ |x|)M/2

1

(1+ |y|)M/2,

since |y| ≤ |ξ |+θ |y−ym| ≤ |ξ |+√

n/N≤ |ξ |+1 for N≥√n. Inserting the estimate

obtained for the integrand in the last displayed integral, we obtain

|DN(x)| ≤C′′

N

|x||α |(1+ |x|)M/2

∫

[−N,N]n

dy

(1+ |y|)M/2+

∫

([−N,N]n)c

|xα∂ βx ϕ(x− y)ψ(y)|dy .


But the second integral in the preceding expression is bounded by

∫

([−N,N]n)c

C′′′|x||α |(1+ |x− y|)M

dy

(1+ |y|)M≤ C′′′|x||α |

(1+ |x|)M/2

∫

([−N,N]n)c

dy

(1+ |y|)M/2.

Using these estimates it is now easy to see that limN→∞ supx∈Rn |DN(x)|= 0.

Next we have the following important result regarding distributions with compact

support:

Theorem 2.3.21. If u is in E ′(Rn), then u is a real analytic function on Rn. In par-

ticular, u is a C ∞ function. Furthermore, u and all of its derivatives have polynomial

growth at infinity. Moreover, u has a holomorphic extension on Cn.

Proof. Given a distribution u with compact support and a polynomial p(ξ ), the ac-

tion of u on the C ∞ function ξ → p(ξ )e−2πix·ξ is a well defined function of x, which

we denote by u(p(·)e−2πix·(·)). Here x is an element of Rn but the same assertion

is valid if x = (x1, . . . ,xn) ∈ Rn is replaced by z = (z1, . . . ,zn) ∈ Cn. In this case we

define the dot product of ξ and z via ξ · z = ∑nk=1 ξkzk.

It is straightforward to verify that the function of z = (z1, . . . ,zn)

F(z) = u(e−2πi(·)·z)

defined on Cn is holomorphic, in fact entire. Indeed, the continuity and linearity of

u and the fact that (e−2πiξ jh− 1)/h→−2πiξ j in C ∞(Rn) as h→ 0, h ∈ C, imply

that F is holomorphic in every variable and its derivative with respect to z j is the

action of the distribution u to the C ∞ function

ξ → (−2πiξ j)e−2πi∑n

j=1 ξ jz j .

By induction it follows that for all multi-indices α we have

∂α1z1· · ·∂αn

znF = u

((−2πi(·))αe

−2πi∑nj=1(·)z j

).

Since F is entire, its restriction on Rn, i.e., F(x1, . . . ,xn), where x j = Re z j, is real

analytic. Also, an easy calculation using (2.3.4) and Leibniz’s rule yield that the

restriction of F on Rn and all of its derivatives have polynomial growth at infinity.

Now for f in S (Rn) we have

⟨u, f

⟩=

⟨u, f

⟩= u

(∫

Rnf (x)e−2πix·ξ dx

)=

∫

Rnf (x)u(e−2πix·(·))dx ,

provided we can justify the passage of u inside the integral. The reason for this

is that the Riemann sums of the integral of f (x)e−2πix·ξ over Rn converge to it in

the topology of C ∞, and thus the linear functional u can be interchanged with the

integral. We conclude that the tempered distribution u can be identified with the real

analytic function x → F(x) whose derivatives have polynomial growth at infinity.

To justify the fact concerning the convergence of the Riemann sums, we argue as

in the proof of the previous theorem. For each N = 1,2, . . . , consider a partition of

[−N,N]n into (2N2)n cubes Qm of side length 1/N and let ym be the center of each

Qm. For a multi-index α let

DN(ξ ) =(2N2)n

∑m=1

f (ym)(−2πiym)αe−2πiym·ξ |Qm|−

∫

Rnf (x)(−2πix)αe−2πix·ξ dx .

We must show that for every M > 0, sup|ξ |≤M |DN(ξ )| converges to zero as N → ∞.

Setting g(x) = f (x)(−2πix)α , we write

DN(ξ ) =(2N2)n

∑m=1

∫

Qm

[g(ym)e

−2πiym·ξ −g(x)e−2πix·ξ ]dx+∫

([−N,N]n)cg(x)e−2πix·ξ dx .

Using the mean value theorem, we bound the absolute value of the expression inside

the square brackets by

(|∇g(zm)|+2π|ξ | |g(zm)|

)√n

N≤ CK (1+ |ξ |)

(1+ |zm|)K

√n

N,

for some point zm in the cube Qm. Since

(2N2)n

∑m=1

∫

Qm

CK (1+ |ξ |)(1+ |zm|)K

dx≤C′K(1+M)< ∞

for |ξ | ≤M, it follows that sup|ξ |≤M |DN(ξ )| → 0 as N → ∞.

Next we give a proposition that extends the properties of the Fourier transform to

tempered distributions.

Proposition 2.3.22. Given u, v in S ′(Rn), f j, f ∈S , y ∈ Rn, b a complex scalar,

α a multi-index, and a > 0, we have

(1) u+ v = u+ v ,

(2) bu = bu ,

(3) If f j → f in S , then f j → f in S and if u j → u in S ′, then u j → u in S ′ ,

(4) (u)= (u) ,

(5) (τyu)= e−2πiy·ξ u ,

(6) (e2πix·yu)= τyu ,

(7) (δ au)= (u)a = a−nδ a−1u ,

(8) (∂αu)= (2πiξ )α u ,

(9) ∂α u = ((−2πix)αu) ,

(10) (u)∨ = u ,

(11) f ∗u = f u ,

(12) f u = f ∗ u ,

(13) (Leibniz’s rule) ∂mj ( f u) = ∑m

k=0

(mk

)(∂ k

j f )(∂m−kj u), m ∈ Z+ ,

(14) (Leibniz’s rule) ∂α( f u) = ∑α1γ1=0 · · ·∑

αnγn=0

(α1γ1

)· · ·

(αn

γn

)(∂ γ f )(∂α−γu) ,

(15) If uk, u ∈ Lp(Rn) and uk → u in Lp (1 ≤ p ≤ ∞), then uk → u in S ′(Rn).Therefore, convergence in S implies convergence in Lp, which in turn implies

convergence in S ′(Rn).

Proof. All the statements can be proved easily using duality and the corresponding

statements for Schwartz functions.

We continue with an application of Theorem 2.3.21.

Proposition 2.3.23. Given u ∈S ′(Rn), there exists a sequence of C ∞0 functions fk

such that fk → u in the sense of tempered distributions; in particular, C ∞0 (Rn) is

dense in S ′(Rn).

Proof. Fix a function in C ∞0 (Rn) with ϕ(x) = 1 in a neighborhood of the origin.

Let ϕk(x) = δ 1/k(ϕ)(x) = ϕ(x/k). It follows from Exercise 2.3.5 (b) that for u ∈S ′(Rn), ϕku → u in S ′. By Proposition 2.3.22 (3), we have that the map u →(ϕku)∨ is continuous on S ′(Rn). Now Theorem 2.3.21 gives that (ϕku)∨ is a C ∞

function and therefore ϕ j(ϕku)∨ is in C ∞0 (Rn). As observed, ϕ j(ϕku)∨→ (ϕku)∨ in

S ′ when k is fixed and j → ∞. Exercise 2.3.5 (c) gives that the diagonal sequence

ϕk(ϕk f )∧ converges to f in S as k→ ∞ for all f ∈S . Using duality and Exercise

2.2.2, we conclude that the sequence of C ∞0 functions ϕk(ϕku)∨ converges to u in

S ′ as k→ ∞.

Exercises

2.3.1. Show that a positive measure µ that satisfies

∫

Rn

dµ(x)

(1+ |x|)k<+∞ ,

for some k > 0, can be identified with a tempered distribution. Show that if we think

of Lebesgue measure as a tempered distribution, then it coincides with the constant

function 1 also interpreted as a tempered distribution.

2.3.2. Let ϕ, f ∈ S (Rn), and for ε > 0 let ϕε(x) = ε−nϕ(ε−1x). Prove that

ϕε ∗ f → b f in S , where b is the integral of ϕ .

2.3.3. Prove that for all a > 0, u ∈S ′(Rn), and f ∈S (Rn) we have

(δ a f )∗ (δ au) = a−nδ a( f ∗u) .

2.3.4. (a) Prove that the derivative of χ[a,b] is δa−δb.

(b) Compute ∂ jχB(0,1) on R2.

(c) Compute the Fourier transforms of the locally integrable functions sinx and cosx.

(d) Prove that the derivative of the distribution log |x| ∈S ′(R) is the distribution

u(ϕ) = limε→0

∫

ε≤|x|

ϕ(x)dx

x.

2.3.5. Let f ∈S (Rn) and let ϕ ∈ C ∞0 be identically equal to 1 in a neighborhood

of the origin. Define ϕk(x) = ϕ(x/k) as in the proof of Proposition 2.3.23.

(a) Prove that (τ−he j f − f )/h→ ∂ j f in S as h→ 0.(b) Prove that ϕk f → f in S as k→ ∞.

(c) Prove that the sequence ϕk(ϕk f )∧ converges to f in S as k→ ∞.

2.3.6. Use Theorem 2.3.21 to show that there does not exist a nonzero C ∞0 function

whose Fourier transform is also a C ∞0 function.

2.3.7. Let f ∈ Lp(Rn) for some 1≤ p < ∞. Show that the sequence of functions

gN(ξ ) =∫

B(0,N)f (x)e−2πix·ξ dx

converges to f in S ′.

2.3.8. Let (ck)k∈Zn be a sequence that satisfies |ck| ≤ A(1+ |k|)M for all k and some

fixed M and A> 0. Let δk denote Dirac mass at the integer k. Show that the sequence

of distributions

∑|k|≤N

ckδk

converges to some tempered distribution u in S ′(Rn) as N → ∞. Also show that u

is the S ′ limit of the sequence of functions

hN(ξ ) = ∑|k|≤N

cke−2πiξ ·k .

2.3.9. A distribution in S ′(Rn) is called homogeneous of degree γ ∈ C if for all

λ > 0 and for all ϕ ∈S (Rn) we have

⟨u,δ λϕ

⟩= λ−n−γ⟨u,ϕ

⟩.

(a) Prove that this definition agrees with the usual definition for functions.

(b) Show that δ0 is homogeneous of degree −n.

(c) Prove that if u is homogeneous of degree γ , then ∂αu is homogeneous of degree

γ−|α|.

2.4 More About Distributions and the Fourier Transform 133

(d) Show that u is homogeneous of degree γ if and only if u is homogeneous of

degree −n− γ .

2.3.10. (a) Show that the functions einx and e−inx converge to zero in S ′ and D ′ as

n→ ∞. Conclude that multiplication of distributions is not a continuous operation

even when it is defined.

(b) What is the limit of√

n(1+n|x|2)−1 in D ′(R) as n→ ∞?

2.3.11. (S. Bernstein) Let f be a bounded function on Rn with f supported in the

ball B(0,R). Prove that for all multi-indices α there exist constants Cα ,n (depending

only on α and on the dimension n) such that

∥∥∂α f∥∥

L∞≤Cα ,nR|α |

∥∥ f∥∥

L∞.

[Hint: Write f = f ∗ h1/R, where h is a Schwartz function h in Rn whose Fourier

transform is equal to one on the ball B(0,1) and vanishes outside the ball B(0,2).]

2.3.12. Let Φ be a C ∞0 function that is equal to 1 in B(0,1) and let Θ be a C ∞

function that is equal to 1 in a neighborhood of infinity and equal to zero in a neigh-

borhood of the origin. Prove the following.

(a) For all u in S ′(Rn) we have

(Φ(ξ/2N

)u)∨→ u in S

′(Rn) as N → ∞.

(b) For all u in S ′(Rn) we have

(Θ(ξ/2N

)u)∨→ 0 in S

′(Rn) as N → ∞.

2.3.13. Prove that there exists a function in Lp for 2 < p < ∞ whose distributional

Fourier transform is not a locally integrable function.[Hint: Assume the converse. Then for all f ∈ Lp(Rn), f is locally integrable and

hence the map f → f is a well defined linear operator from Lp(Rn) to L1(B(0,M))

for all M > 0 (i.e. ‖ f ‖L1(B(0,M)) < ∞ for all f ∈ Lp(Rn)). Use the closed graph

theorem to deduce that ‖ f ‖L1(B(0,M)) ≤CM‖ f‖Lp(Rn) for some CM < ∞. To violate

this inequality whenever p > 2, take fN(x) = (1 + iN)−n/2e−π(1+iN)−1|x|2 and let

N → ∞, noting that fN(ξ ) = e−π|ξ |2(1+iN).

]

2.4 More About Distributions and the Fourier Transform

In this section we discuss further properties of distributions and Fourier transforms

and bring up certain connections that arise between harmonic analysis and partial

differential equations.


2.4.1 Distributions Supported at a Point

We begin with the following characterization of distributions supported at a single

point.

Proposition 2.4.1. If u∈S ′(Rn) is supported in the singleton x0, then there exists

an integer k and complex numbers aα such that

u = ∑|α |≤k

aα∂αδx0

.

Proof. Without loss of generality we may assume that x0 = 0. By (2.3.3) we have

that for some C, m, and k,

∣∣⟨u, f⟩∣∣≤C ∑

|α |≤m|β |≤k

supx∈Rn

|xα(∂ β f )(x)| for all f ∈S (Rn).

We now prove that if ϕ ∈S satisfies

(∂αϕ)(0) = 0 for all |α| ≤ k, (2.4.1)

then⟨u,ϕ

⟩= 0. To see this, fix a ϕ satisfying (2.4.1) and let ζ (x) be a smooth

function on Rn that is equal to 1 when |x| ≥ 2 and equal to zero for |x| ≤ 1. Let

ζ ε(x) = ζ (x/ε). Then, using (2.4.1) and the continuity of the derivatives of ϕ at the

origin, it is not hard to show that ρα ,β (ζεϕ−ϕ)→ 0 as ε → 0 for all |α| ≤ m and

|β | ≤ k. Then

∣∣⟨u,ϕ⟩∣∣≤

∣∣⟨u,ζ εϕ⟩∣∣+

∣∣⟨u,ζ εϕ−ϕ⟩∣∣≤ 0+C ∑

|α |≤m|β |≤k

ρα ,β (ζεϕ−ϕ)→ 0

as ε → 0. This proves our assertion.

Now let f ∈S (Rn). Let η be a C ∞0 function on Rn that is equal to 1 in a neigh-

borhood of the origin. Write

f (x) = η(x)(∑|α |≤k

(∂α f )(0)

α!xα +h(x)

)+(1−η(x)) f (x), (2.4.2)

where h(x) = O(xk+1) as |x| → 0. Then ηh satisfies (2.4.1) and hence⟨u,ηh

⟩= 0

by the claim. Also, ⟨u,((1−η) f

)⟩= 0

by our hypothesis. Applying u to both sides of (2.4.2), we obtain

⟨u, f

⟩= ∑|α |≤k

(∂α f )(0)

α!u(xαη(x)) = ∑

|α |≤k

aα(∂αδ0)( f ) ,

with aα = (−1)|α |u(xαη(x))/α!. This proves the proposition.


An immediate consequence is the following result.

Corollary 2.4.2. Let u ∈S ′(Rn). If u is supported in the singleton ξ0, then u is

a finite linear combination of functions (−2πiξ )αe2πiξ ·ξ0 , where α is a multi-index.

In particular, if u is supported at the origin, then u is a polynomial.

Proof. Proposition 2.4.1 gives that u is a linear combination of derivatives of Dirac

masses at ξ0. Then Proposition 2.3.22 (8) yields the required conclusion.

2.4.2 The Laplacian

The Laplacian ∆ is a partial differential operator acting on tempered distributions

on Rn as follows:

∆(u) =n

∑j=1

∂ 2j u .

Solutions of Laplace’s equation ∆(u)= 0 are called harmonic distributions. We have

the following:

Corollary 2.4.3. Let u ∈S ′(Rn) satisfy ∆(u) = 0. Then u is a polynomial.

Proof. Taking Fourier transforms, we obtain that ∆(u) = 0. Therefore,

−4π2|ξ |2u = 0 in S′.

This implies that u is supported at the origin, and by Corollary 2.4.2 it follows that

u must be polynomial.

Liouville’s classical theorem that every bounded harmonic function must be con-

stant is a consequence of Corollary 2.4.3. See Exercise 2.4.2.

Next we would like to compute the fundamental solutions of Laplace’s equation

in Rn. A distribution is called a fundamental solution of a partial differential operator

L if we have L(u) = δ0. The following result gives the fundamental solution of the

Laplacian.

Proposition 2.4.4. For n≥ 3 we have

∆(|x|2−n) =−(n−2)2πn/2

Γ (n/2)δ0 , (2.4.3)

while for n = 2,

∆(log |x|) = 2πδ0 . (2.4.4)

Proof. We use Green’s identity

∫

Ω

(v∆(u)−u∆(v)

)dx =

∫

∂Ω

(v∂u

∂ν−u

∂v

∂ν

)ds ,


where Ω is an open set in Rn with smooth boundary and ∂v/∂ν denotes the

derivative of v with respect to the outer unit normal vector. Take Ω = Rn \B(0,ε),v= |x|2−n, and u= f a C ∞

0 (Rn) function in the previous identity. The normal deriva-

tive of f (rθ) is the derivative with respect to the radial variable r. Observe that

∆(|x|2−n) = 0 for x = 0. We obtain

∫

|x|>ε∆( f )(x)|x|2−n dx =−

∫

|θ |=ε

(ε2−n ∂ f

∂ r− f (rθ)

∂ r2−n

∂ r

)dθ , (2.4.5)

where dθ denotes surface measure on the sphere |θ |= ε . Now observe two things:

first, that for some C =C( f ) we have

∣∣∣∣∫

|θ |=ε

∂ f

∂ rdθ

∣∣∣∣≤Cεn−1 ;

second, that ∫

|θ |=εf (rθ)ε1−n dθ → ωn−1 f (0)

as ε → 0. Letting ε → 0 in (2.4.5), we obtain that

limε→0

∫

|x|>ε∆( f )(x)|x|2−n dx =−(n−2)ωn−1 f (0) ,

which implies (2.4.3) in view of the formula for ωn−1 given in Appendix A.3.

The proof of (2.4.4) is identical. The only difference is that the quantity ∂ r2−n/∂ r

in (2.4.5) is replaced by ∂ logr/∂ r.

2.4.3 Homogeneous Distributions

The fundamental solutions of the Laplacian are locally integrable functions on Rn

and also homogeneous of degree 2− n when n ≥ 3. Since homogeneous distribu-

tions often arise in applications, it is desirable to pursue their study. Here we do not

undertake such a study in depth, but we discuss a few important examples.

Our first goal is to understand the action of the distribution |t|z on Rn when

Rez≤−n. Let us consider first the case n = 1. The tempered distribution

〈wz,ϕ〉=∫ 1

−1|t|zϕ(t)dt

is well-defined when Rez > −1. But we can extend the definition for all z with

Rez >−3 and z =−1 by rewriting it as

〈wz,ϕ〉=∫ 1

−1|t|z

(ϕ(t)−ϕ(0)− tϕ ′(0)

)dt +

2

z+1ϕ(0) , (2.4.6)

and noting that for all ϕ ∈S (R) we have

∣∣〈wz,ϕ〉∣∣≤ 1

z+3‖ϕ ′′‖L∞ +

2

z+1‖ϕ‖L∞ ,

thus wz ∈ S ′(R). Subtracting the Taylor polynomial of degree 3 centered at zero

from ϕ(t) instead of the linear one, as in (2.4.6), allows us to extend the definition

for Rez > −5 and Rez /∈ −1,−3. Subtracting higher order Taylor polynomials

allows us to extend the definition of wz for all z ∈ C except at the negative odd

integers. To be able to include the points z=−1,−3,−5,−7, . . . we need to multiply

wz by an entire function that has simple zeros at all the negative odd integers to be

able to eliminate the simple poles at these points. Such a function is Γ(

z+12

)−1. This

discussion leads to the following definition.

Definition 2.4.5. For z ∈ C we define a distribution uz as follows:

⟨uz, f

⟩=

∫

Rn

πz+n

2

Γ(

z+n2

) |x|z f (x)dx . (2.4.7)

Clearly the uz’s coincide with the locally integrable functions

πz+n

2 Γ(

z+n2

)−1|x|z

when Rez >−n and the definition makes sense only for that range of z’s. It follows

from its definition that uz is a homogeneous distribution of degree z.

We would like to extend the definition of uz for z ∈ C. Let Re z >−n first. Fix N

to be a positive integer. Given f ∈S (Rn), write the integral in (2.4.7) as follows:

∫

|x|<1

πz+n

2

Γ ( z+n2)

f (x)− ∑

|α |≤N

(∂α f )(0)

α!xα

|x|z dx

+∫

|x|>1

πz+n

2

Γ ( z+n2)

f (x)|x|z dx+∫

|x|<1

πz+n

2

Γ ( z+n2)∑|α |≤N

(∂α f )(0)

α!xα |x|z dx .

The preceding expression is equal to

∫

|x|<1

πz+n

2

Γ ( z+n2)

f (x)− ∑

|α |≤N

(∂α f )(0)

α!xα

|x|z dx

+∫

|x|>1

πz+n

2

Γ ( z+n2)

f (x)|x|z dx

+ ∑|α |≤N

(∂α f )(0)

α!

πz+n

2

Γ ( z+n2)

∫ 1

r=0

∫

Sn−1(rθ)α rz+n−1 dr dθ ,

where we switched to polar coordinates in the penultimate integral. Now set

b(n,α,z) =π

z+n2

Γ ( z+n2)

1

α!

(∫

Sn−1θα dθ

)∫ 1

0r|α |+n+z−1 dr

=π

z+n2

Γ ( z+n2)

1α!

∫

Sn−1θα dθ

|α|+ z+n,

where α = (α1, . . . ,αn) is a multi-index. These coefficients are zero when at least

one α j is odd. Consider now the case that all the α j’s are even; then |α| is also even.

The function Γ ( z+n2) has simple poles at the points

z =−n, z =−(n+2), z =−(n+4), and so on;

see Appendix A.5. These poles cancel exactly the poles of the function

z → (|α|+ z+n)−1

at z =−n−|α| when |α| is an even integer in [0,N]. We therefore have

⟨uz, f

⟩=

∫

|x|≥1

πz+n

2

Γ ( z+n2)

f (x)|x|z dx+ ∑|α |≤N

b(n,α,z)(−1)|α |⟨∂αδ0, f

⟩

+∫

|x|<1

πz+n

2

Γ ( z+n2)

f (x)− ∑

|α |≤N

(∂α f )(0)

α!xα

|x|z dx .

(2.4.8)

Both integrals converge absolutely when Re z > −N− n− 1, since the expression

inside the curly brackets above is bounded by a constant multiple of |x|N+1, and

the resulting function of z in (2.4.8) is a well defined analytic function in the range

Re z >−N−n−1.

Since N was arbitrary, 〈uz, f 〉 has an analytic extension to all of C. Therefore,

uz is a distribution-valued entire function of z, i.e., for all ϕ ∈S (Rn), the function

z → 〈uz,ϕ〉 is entire.

Next we would like to calculate the Fourier transform of uz. We know by Exercise

2.3.9 that uz is a homogeneous distribution of degree−n− z. The choice of constant

in the definition of uz was made to justify the following result:

Theorem 2.4.6. For all z ∈ C we have uz = u−n−z.

Proof. The idea of the proof is straightforward. First we show that for a certain range

of z’s we have

∫

Rn|ξ |zϕ(ξ )dξ =C(n,z)

∫

Rn|x|−n−zϕ(x)dx , (2.4.9)

for some fixed constant C(n,z) and all ϕ ∈ S (Rn). Next we pick a specific ϕ to

evaluate the constant C(n,z). Then we use analytic continuation to extend the va-

lidity of (2.4.9) for all z’s. Use polar coordinates by setting ξ = ρϕ and x = rθ in

(2.4.9). We have

∫

Rn|ξ |zϕ(ξ )dξ

=∫ ∞

0ρz+n−1

∫ ∞

0

∫

Sn−1ϕ(rθ)

(∫

Sn−1e−2πirρ(θ ·ϕ)dϕ

)dθ rn−1 dr dρ

=∫ ∞

0

(∫ ∞

0σn(rρ)ρ

z+n−1 dρ

)(∫

Sn−1ϕ(rθ)dθ

)rn−1 dr

=C(n,z)∫ ∞

0r−z−n

(∫

Sn−1ϕ(rθ)dθ

)rn−1 dr

=C(n,z)∫

Rn|x|−n−zϕ(x)dx ,

where we set

σn(t) =∫

Sn−1e−2πit(θ ·ϕ) dϕ =

∫

Sn−1e−2πit(ϕ1) dϕ , (2.4.10)

C(n,z) =

∫ ∞

0σn(t)t

z+n−1 dt , (2.4.11)

and the second equality in (2.4.10) is a consequence of rotational invariance. It re-

mains to prove that the integral in (2.4.11) converges for some range of z’s.

If n = 1, then

σ1(t) =∫

S0e−2πitϕdϕ = e−2πit + e2πit = 2cos(2πt)

and the integral in (2.4.11) converges conditionally for −1 < Re z < 0.

Let us therefore assume that n≥ 2. Since |σn(t)| ≤ ωn−1, the integral converges

near zero when −n < Re z. Let us study the behavior of σn(t) for t large. Using the

formula in Appendix D.2 and the definition of Bessel functions in Appendix B.1, we

write

σn(t) = ωn−2

∫ 1

−1e2πits

(√1− s2

)n−2 ds√1− s2

= cn t−n−2

2 J n−22(2πt),

for some constant cn. Since n≥ 2 we have when n−2>−1/2. Then the asymptotics

for Bessel functions (Appendix B.7) apply and yield |σn(t)| ≤ ct−(n−1)/2 for t ≥ 1.

Splitting the integral in (2.4.11) in t ≤ 1 and t ≥ 1 and using the corresponding

estimates, we notice that it converges absolutely on [0,1] when Rez > −n and on

[1,∞) when Re z+n−1− n−12

<−1.

We have now proved that when −n < Re z <− n+12

and n≥ 2 we have

uz =C(n,z)u−n−z

for some constant C(n,z) that we wish to compute. Insert the function ϕ(x) = e−π|x|2

in (2.4.9). Example 2.2.9 gives that this function is equal to its Fourier transform.

Use polar coordinates to write

ωn−1

∫ ∞

0rz+n−1e−πr2

dr =C(n,z)ωn−1

∫ ∞

0r−z−n+n−1e−πr2

dr .

Change variables s = πr2 and use the definition of the gamma function to obtain

that

C(n,z) =Γ ( z+n

2)

Γ (− z2)

π−z+n

2

πz2

.

It follows that uz = u−n−z for the range of z’s considered.

At this point observe that for every f ∈S (Rn), the function z → 〈uz−u−z−n, f 〉is entire and vanishes for −n < Rez < −n+ 1/2. Therefore, it must vanish every-

where and the theorem is proved.

Homogeneous distributions were introduced in Exercise 2.3.9. We already saw

that the Dirac mass on Rn is a homogeneous distribution of degree −n. There is

another important example of a homogeneous distributions of degree −n, which we

now discuss.

Let Ω be an integrable function on the sphere Sn−1 with integral zero. Define a

tempered distribution WΩ on Rn by setting

⟨WΩ , f

⟩= lim

ε→0

∫

|x|≥ε

Ω(x/|x|)|x|n f (x)dx . (2.4.12)

We check that WΩ is a well defined tempered distribution on Rn. Indeed, since

Ω(x/|x|)/|x|n has integral zero over all annuli centered at the origin, we obtain

∣∣⟨WΩ ,ϕ⟩∣∣ =

∣∣∣∣limε→0

∫

ε≤|x|≤1

Ω(x/|x|)|x|n (ϕ(x)−ϕ(0))dx+

∫

|x|≥1

Ω(x/|x|)|x|n ϕ(x)dx

∣∣∣∣

≤∥∥∇ϕ

∥∥L∞

∫

|x|≤1

|Ω(x/|x|)||x|n−1

dx+

(supx∈Rn

|x| |ϕ(x)|)∫

|x|≥1

|Ω(x/|x|)||x|n+1

dx

≤C1

∥∥∇ϕ∥∥

L∞

∥∥Ω∥∥

L1(Sn−1)+C2 ∑

|α |≤1

∥∥ϕ(x)xα∥∥

L∞

∥∥Ω∥∥

L1(Sn−1),

for suitable constants C1 and C2 in view of (2.2.2).

One can verify that WΩ ∈S ′(Rn) is a homogeneous distribution of degree −n

just like the Dirac mass at the origin. It is an interesting fact that all homogeneous

distributions on Rn of degree −n that coincide with a smooth function away from

the origin arise in this way. We have the following result.

Proposition 2.4.7. Suppose that m is a C ∞ function on Rn\0 that is homogeneous

of degree zero. Then there exist a scalar b and a C ∞ functionΩ on Sn−1 with integral

zero such that

m∨ = bδ0 +WΩ , (2.4.13)

where WΩ denotes the distribution defined in (2.4.12).

To prove this result we need the following proposition, whose proof we postpone

until the end of this section.

Proposition 2.4.8. Suppose that u is a C ∞ function on Rn \0 that is homogeneous

of degree z ∈ C. Then u is a C ∞ function on Rn \0.

We now prove Proposition 2.4.7 using Proposition 2.4.8.

Proof. Let a be the integral of the smooth function m over Sn−1. The function m−a

is homogeneous of degree zero and thus locally integrable on Rn; hence it can be

thought of as a tempered distribution that we call u (the Fourier transform of a

tempered distribution u). Since u is a C ∞ function on Rn \ 0, Proposition 2.4.8

implies that u is also a C ∞ function on Rn \ 0. Let Ω be the restriction of u on

Sn−1. Then Ω is a well defined C ∞ function on Sn−1. Since u is a homogeneous

function of degree−n that coincides with the smooth function Ω on Sn−1, it follows

that u(x) =Ω(x/|x|)/|x|n for x in Rn \0.We show that Ω has mean value zero over Sn−1. Pick a nonnegative, radial,

smooth, and nonzero function ψ on Rn supported in the annulus 1< |x|< 2. Switch-

ing to polar coordinates, we write

⟨u,ψ

⟩=

∫

Rn

Ω(x/|x|)|x|n ψ(x)dx = cψ

∫

Sn−1Ω(θ)dθ ,

⟨u,ψ

⟩=

⟨u, ψ

⟩=

∫

Rn(m(ξ )−a)ψ(ξ )dξ = c′ψ

∫

Sn−1

(m(θ)−a

)dθ = 0 ,

and thus Ω has mean value zero over Sn−1 (since cψ = 0).

We can now legitimately define the distribution WΩ , which coincides with the

function Ω(x/|x|)/|x|n on Rn \ 0. But the distribution u also coincides with this

function on Rn \ 0. It follows that u−WΩ is supported at the origin. Proposition

2.4.1 now gives that u−WΩ is a sum of derivatives of Dirac masses. Since both

distributions are homogeneous of degree −n, it follows that

u−WΩ = cδ0 .

But u= (m−a)∨ = m∨−aδ0, and thus m∨ = (c+a)δ0+WΩ . This proves the propo-

sition.

We now turn to the proof of Proposition 2.4.8.

Proof. Let u ∈S ′ be homogeneous of degree z and C ∞ on Rn \ 0. We need to

show that u is C ∞ away from the origin. We prove that u is C M for all M. Fix M ∈Z+

and let α be any multi-index such that

|α|> n+M+Re z . (2.4.14)

Pick a C ∞ function ϕ on Rn that is equal to 1 when |x| ≥ 2 and equal to zero for

|x| ≤ 1. Write u0 = (1−ϕ)u and u∞ = ϕu. Then

∂αu = ∂αu0 +∂αu∞ and thus ∂αu = ∂αu0 + ∂αu∞ ,

where the operations are performed in the sense of distributions. Since u0 is com-

pactly supported, Theorem 2.3.21 implies that ∂αu0 is C ∞. Now Leibniz’s rule gives

that

∂αu∞ = v+ϕ∂αu,

where v is a smooth function supported in the annulus 1 ≤ |x| ≤ 2. Then v is

C ∞ and we need to show only that ϕ∂αu is C M . The function ϕ∂αu is ac-

tually C ∞, and by the homogeneity of ∂αu (Exercise 2.3.9 (c)) we obtain that

(∂αu)(x) = |x|−|α |+z(∂αu)(x/|x|). Since ϕ is supported away from zero, it follows

that

|ϕ(x)(∂αu)(x)| ≤ Cα

(1+ |x|)|α |−Re z(2.4.15)

for some Cα > 0. It is now straightforward to see that if a function satisfies (2.4.15),

then its Fourier transform is C M whenever (2.4.14) is satisfied. See Exercise 2.4.1.

We conclude that ∂αu∞ is a C M function whenever (2.4.14) is satisfied; thus so

is ∂αu. Since ∂αu(ξ ) = (2πiξ )α u(ξ ), we deduce smoothness for u away from the

origin. Let ξ = 0. Pick a neighborhood V of ξ such that for η in V we have η j = 0

for some j ∈ 1, . . . ,n. Consider the multi-index (0, . . . , |α|, . . . ,0) with |α| in the

jth coordinate and zeros elsewhere. Then (2πiη j)|α |u(η) is a C M function on V ,

and thus so is u(η), since we can divide by η|α |j . We conclude that u(ξ ) is C M on

Rn \0. Since M is arbitrary, the conclusion follows.

We end this section with an example that illustrates the usefulness of some of the

ideas discussed in this section.

Example 2.4.9. Let η be a smooth radial function on Rn that is equal to 1 on the

set |x| ≥ 1/2 and vanishes on the set |x| ≤ 1/4. Fix z ∈ C satisfy 0 < Rez < n. Let

g =(η(x)|x|−z

)∧be the distributional Fourier transform of η(x)|x|−z. We show that

g is a function that decays faster than |ξ |−N at infinity (for sufficiently large positive

number N) and that

g(ξ )− πz− n2Γ ( n−z

2)

Γ ( z2)

|ξ |z−n (2.4.16)

is a C ∞ function on Rn. This example indicates the interplay between the smooth-

ness of a function and the decay of its Fourier transform. The smoothness of the

function η(x)|x|−z near zero has as a consequence the rapid decay of g near infinity,

while the slow decay of η(x)|x|−z at infinity reflects the lack of smoothness of g(ξ )at zero, in view of the moderate blowup |ξ |Rez−n as |ξ | → 0.

To show that g is a function we write it as g = (|x|−z)∧+((η(x)−1)|x|−z

)∧and

we observe that the first term is a function, since 0 < Rez < n. Using Theorem 2.4.6

we write

g(ξ ) =πz− n

2Γ ( n−z2)

Γ ( z2)

|ξ |z−n + ϕ(ξ ) ,

where ϕ(ξ ) =((η(x)−1)|x|−z

)∧(ξ ) is a C ∞ function, since it is the Fourier trans-

form of a compactly supported integrable function. This proves that g is a function

and that the difference in (2.4.16) is C ∞.

Finally, we assert that every derivative of g satisfies |∂ γg(ξ )| ≤Cγ ,N |ξ |−N for all

sufficiently large positive integers N when ξ = 0. Indeed, fix a multi-index γ and

write ∂ γg(ξ ) = (|x|−zη(x)(−2πix)γ)∧(ξ ). It follows that

(4π2|ξ |2)N |∂ γg(ξ )|=∣∣(∆N(|x|−zη(x)(−2πix)γ)

)∧(ξ )

∣∣

for all N ∈ Z+, where ∆ is the Laplacian in the x variable. Using Leibniz’s rule

we distribute ∆N to the product. If a derivative falls on η , we obtain a compactly

supported smooth function, hence integrable. If all derivatives fall on |x|−zxγ , then

we obtain a term that decays like |x|−Rez+|γ |−2N at infinity, which is also integrable

if N is sufficiently large. Thus the function |ξ |2N |∂ γg(ξ )| is equal to the Fourier

transform of an L1 function, hence it is bounded, when 2N > n−Rez+ |γ |.

Exercises

2.4.1. Suppose that a function f satisfies the estimate

| f (x)| ≤ C

(1+ |x|)N,

for some C > 0 and N > n+1. Then f is C M for all M ∈ Z+ with 1≤M < N−n.

2.4.2. Use Corollary 2.4.3 to prove Liouville’s theorem that every bounded har-

monic function on Rn must be a constant. Derive as a consequence the fundamental

theorem of algebra, stating that every polynomial on C must have a complex root.

2.4.3. Prove that ex is not in S ′(R) but that exeiexis in S ′(R).

2.4.4. Show that the Schwartz function x → sech(πx), x ∈ R, coincides with its

Fourier transform.[Hint: Integrate the function eiaz over the rectangular contour with corners (−R,0),(R,0), (R, iπ), and (−R, iπ).

]

2.4.5. ([174]) Construct an uncountable family of linearly independent Schwartz

functions fa such that | fa|= | fb| and | fa|= | fb| for all fa and fb in the family.[Hint: Let w be a smooth nonzero function whose Fourier transform is supported

in the interval [−1/2,1/2] and let φ be a real-valued smooth nonconstant periodic

function with period 1. Then take fa(x) = w(x)eiφ(x−a) for a ∈ R.]

2.4.6. Let Py be the Poisson kernel defined in (2.1.13). Prove that for f ∈ Lp(Rn),1≤ p < ∞, the function

(x,y) → (Py ∗ f )(x)

is a harmonic function on Rn+1+ . Use the Fourier transform and Exercise 2.2.11 to

prove that (Py1∗Py2

)(x) = Py1+y2(x) for all x ∈ Rn.

2.4.7. (a) For a fixed x0 ∈ Sn−1, show that the function

v(x;x0) =1−|x|2|x− x0|n

is harmonic on Rn \x0.(b) For fixed x0 ∈ Sn−1, prove that the family of functions θ → v(rx0;θ), 0 < r < 1,

defined on the sphere satisfies

limr↑1

∫

θ∈Sn−1

|θ−x0|>δv(rx0;θ)dθ = 0

uniformly in x0. The function v(rx0;θ) is called the Poisson kernel for the sphere.

(c) Show that1

ωn−1(1−|x|2)

∫

Sn−1

1

|x−θ |n dθ = 1

for all |x|< 1.

(d) Let f be a continuous function on Sn−1. Prove that the function

u(x) =1

ωn−1(1−|x|2)

∫

Sn−1

f (θ)

|x−θ |n dθ

solves the Dirichlet problem ∆(u) = 0 on |x| < 1 with boundary values u = f on

Sn−1, in the sense limr↑1 u(rx0) = f (x0) when |x0|= 1.[Hint: Part (c): Apply the mean value property over spheres to the harmonic function

y → (1−|x|2|y|2)∣∣|x|y− x

|x|∣∣−n

.]

2.4.8. Fix n ∈ Z+ with n≥ 2 and a real number λ , 0 < λ < n. Also fix η ∈ Sn and

y ∈ Rn.

(a) Prove that∫

Sn|ξ −η |−λ dξ = 2n−λ π

n2Γ ( n−λ

2)

Γ (n− λ2).

(b) Prove that

∫

Rn|x− y|−λ (1+ |x|2) λ2−n dx =

πn2Γ ( n−λ

2)

Γ (n− λ2)(1+ |y|2)− λ

2 .

[Hint: Part (a): See Appendix D.4 Part (b): Use the stereographic projection in

Appendix D.6.]

2.4.9. Prove the following beta integral identity:

∫

Rn

dt

|x− t|α1 |y− t|α2= π

n2Γ(

n−α12

)Γ(

n−α22

)Γ(α1+α2−n

2

)

Γ(α1

2

)Γ(α2

2

)Γ(n− α1+α2

2

) |x− y|n−α1−α2 ,

where 0 < α1,α2 < n, α1 +α2 > n.[Hint: Reduce to the case y = 0, interpret the integral as a convolution, and use

Theorem 2.4.6.]

2.4.10. (a) Prove that if a continuous integrable function f on Rn (n≥ 2) is constant

on the spheres r Sn−1 for all r > 0, then so is its Fourier transform.

(b) If a continuous integrable function on Rn (n ≥ 3) is constant on all (n− 2)–dimensional spheres orthogonal to e1 = (1,0, . . . ,0), then its Fourier transform has

the same property.

2.4.11. ([137]) Suppose that 0 < d1,d2,d3 < n satisfy d1 +d2 +d3 = 2n. Prove that

for any distinct x,y,z ∈ Rn we have the identity

∫

Rn|x− t|−d2 |y− t|−d3 |z− t|−d1dt

= πn2

(3

∏j=1

Γ(n− d j

2

)

Γ( d j

2

))|x− y|d1−n|y− z|d2−n|z− x|d3−n.

[Hint: Reduce matters to the case that z = 0 and y = e1. Then take the Fourier

transform in x and use that the function h(t) = |t − e1|−d3 |t|−d1 satisfies h(ξ ) =

h(A−2ξξ ) for all ξ = 0, where Aξ is an orthogonal matrix with Aξ e1 = ξ/|ξ |.

]

2.4.12. (a) Integrate the function eiz2over the contour consisting of the three pieces

P1 = x+ i0 : 0≤ x≤ R, P2 = Reiθ : 0≤ θ ≤ π4, and P3 = r ei π4 : 0≤ r ≤ R

(with the proper orientation) to obtain the Fresnel integral identity:

limR→∞

∫ R

0eix2

dx =√

2π4

(1+ i) .

(b) Use the result in part (a) to show that the Fourier transform of the function eiπ|x|2

in Rn is equal to ei πn4 e−iπ|ξ |2 .[

Hint: Part (a): On P2 we have e−R2sin(2θ) ≤ e−4π R2θ , and the integral over P2 tends

to 0. Part (b): Try first n = 1.]


2.5 Convolution Operators on Lp Spaces and Multipliers

In this section we study the class of operators that commute with translations. We

prove in this section that bounded operators that commute with translations must be

of convolution type. Convolution operators arise in many situations, and we would

like to know under what circumstances they are bounded between Lp spaces.

2.5.1 Operators That Commute with Translations

Definition 2.5.1. A vector space X of measurable functions on Rn is called closed

under translations if for f ∈ X we have τz( f ) ∈ X for all z ∈ Rn. Let X and Y be

vector spaces of measurable functions on Rn that are closed under translations. Let

also T be an operator from X to Y . We say that T commutes with translations or is

translation-invariant if

T (τy( f )) = τy(T ( f ))

for all f ∈ X and all y ∈ Rn.

It is straightforward to see that convolution operators commute with translations,

i.e., τy( f ∗ g) = τy( f ) ∗ g whenever the convolution is defined. One of the goals of

this section is to prove the converse: every bounded linear operator that commutes

with translations is of convolution type. We have the following:

Theorem 2.5.2. Let 1≤ p,q≤ ∞ and suppose T is a bounded linear operator from

Lp(Rn) to Lq(Rn) that commutes with translations. Then there exists a unique tem-

pered distribution w such that

T ( f ) = f ∗w a.e. for all f ∈S .

A very important point to make is that if p = ∞, the restriction of T on S does

not uniquely determine T on the entire L∞; see Example 2.5.9 and the comments

preceding it about this. The theorem is a consequence of the following two results:

Lemma 2.5.3. Under the hypotheses of Theorem 2.5.2 and for f ∈S (Rn), the dis-

tributional derivatives of T ( f ) are Lq functions that satisfy

∂α(T ( f )

)= T (∂α f ), for all multi-indices α . (2.5.1)

Lemma 2.5.4. Let 1 ≤ q ≤ ∞ and let h ∈ Lq(Rn). If all distributional derivatives

∂αh are also in Lq, then h is almost everywhere equal to a continuous function H

satisfying

|H(0)| ≤Cn,q ∑|α |≤n+1

∥∥∂αh∥∥

Lq . (2.5.2)

2.5 Convolution Operators on Lp Spaces and Multipliers 147

Proof. Assuming Lemmas 2.5.3 and 2.5.4, we prove Theorem 2.5.2.

Given f ∈S (Rn), by Lemmas 2.5.3 and 2.5.4, there is a continuous function H

such that T ( f ) = H a.e. and such that

|H(0)| ≤Cn,q ∑|α |≤n+1

∥∥∂αT ( f )∥∥

Lq

holds. Define a linear functional u on S by setting

⟨u, f

⟩= H(0).

This functional is well-defined, for, if there is another continuous function G such

that G = T ( f ) a.e., then G = H a.e. and since both functions are continuous, it

follows that H = G everywhere and thus H(0) = G(0).By (2.5.1), (2.5.2), and the boundedness of T , we have

∣∣⟨u, f⟩∣∣≤ Cn,q ∑

|α |≤n+1

∥∥∂αT ( f )∥∥

Lq

≤Cn,q ∑|α |≤n+1

∥∥T (∂α f )∥∥

Lq

≤Cn,q

∥∥T∥∥

Lp→Lq ∑|α |≤n+1

∥∥∂α f∥∥

Lp

≤C′n,q∥∥T

∥∥Lp→Lq ∑

|γ |≤[ n+1p ]+1

|α |≤n+1

ργ ,α( f ) ,

where the last estimate uses (2.2.8). This implies that u is in S ′. We now set w = u

and we claim that for all x ∈ Rn we have

⟨u,τ−x f

⟩= H(x) . (2.5.3)

Assuming (2.5.3) we prove that T ( f ) = f ∗w for f ∈S . To see this, by Theorem

2.3.20 and by the translation invariance of T , for a given f ∈S (Rn) we have

( f ∗w)(x) =⟨u,τx f

⟩=

⟨u,τ−x f

⟩= H(x) = T ( f )(x) ,

where the last equality holds for almost all x, by the definition of H. Thus f ∗w =T ( f ) a.e., as claimed. The uniqueness of w follows from the simple observation that

if f ∗w = f ∗w′ for all f ∈S (Rn), then w = w′.We now turn to the proof of (2.5.3). Given f ∈ S (Rn) and x ∈ Rn and let Hx

be the continuous function such that Hx = T (τ−x f ). We show that Hx(0) = H(x).Indeed, we have

Hx(y) = T (τ−x f )(y) = τ−xT ( f )(y) = T ( f )(x+ y) = H(x+ y) = τ−xH(y) ,

where the equality T ( f )(x + y) = H(x + y) holds a.e. in y. Thus the continuous

functions Hx and τ−xH are equal a.e. and thus they must be everywhere equal,

in particular, when y = 0. This proves that Hx(0) = H(x), which is a restatement

of (2.5.3).

We now return to Lemmas 2.5.3 and 2.5.4. We begin with Lemma 2.5.3.

Proof. Consider first the multi-index α = (0, . . . ,1, . . . ,0), where 1 is in the jth

entry and 0 is elsewhere. Let e j = (0, . . . ,1, . . . ,0), where 1 is in the jth entry and

zero elsewhere. We have

∫

RnT ( f )(y)

ϕ(y+he j)−ϕ(y)h

dy =∫

Rnϕ(y)T

(τhe j( f )− f

h

)(y)dy (2.5.4)

since both of these expressions are equal to

∫

Rnϕ(y)

T ( f )(y−he j)−T ( f )(y)

hdy

and T commutes with translations. We will let h → 0 in both sides of (2.5.4). We

writeϕ(y+he j)−ϕ(y)

h=

∫ 1

0∂ jϕ(y+hte j)dt ,

from which it follows that for |h|< 1/2 we have

∣∣∣∣ϕ(y+he j)−ϕ(y)

h

∣∣∣∣≤∫ 1

0

CM dt

(1+ |y+hte j|)M≤

∫ 1

0

CM dt

(1+ |y|− 12)M≤ C′M

(|y|+1)M.

The integrand on the left-hand side of (2.5.4) is bounded by the integrable function

|T ( f )(y)|C′M(|y|+ 1)−M and converges to T ( f )(y)∂ jϕ(y) as h→ 0. The Lebesgue

dominated convergence theorem yields that the integral on the left-hand side of

(2.5.4) converges to ∫

RnT ( f )(y)∂ jϕ(y)dy . (2.5.5)

Moreover, for a Schwartz function f we have

τhe j( f )(y)− f (y)

h=

∫ 1

0∂ j f (y+hte j)dt ,

which converges to ∂ j f (y) pointwise as h→ 0 and is bounded by C′M(1+ |y|)−M for

|h|< 1/2 by an argument similar to the preceding one for ϕ in place of f . Thus

τhe j( f )− f

h→ ∂ j f in Lp as h→ 0, (2.5.6)

by the Lebesgue dominated convergence theorem. The boundedness of T from Lp

to Lq yields that

T

(τhe j( f )− f

h

)→ T (∂ j f ) in Lq as h→ 0. (2.5.7)

Since ϕ ∈ Lq′ , by Holder’s inequality, the right-hand side of (2.5.4) converges to

∫

Rnϕ(y)T (∂ j f )(y)dy

as h→ 0. This limit is equal to (2.5.5) and the required conclusion follows for α =(0, . . . ,0,1,0, . . . ,0). The general case follows by induction on |α|.

We now prove Lemma 2.5.4.

Proof. Let R ≥ 1. Fix a C ∞0 function ϕR that is equal to 1 in the ball |x| ≤ R and

equal to zero when |x| ≥ 2R. Since h is in Lq(Rn), it follows that ϕRh is in L1(Rn).

We show that ϕRh is also in L1. We begin with the inequality

1≤Cn(1+ |x|)−(n+1) ∑|α |≤n+1

|(−2πix)α | , (2.5.8)

which is just a restatement of (2.2.3). Now multiply (2.5.8) by |ϕRh(x)| to obtain

|ϕRh(x)| ≤Cn(1+ |x|)−(n+1) ∑|α |≤n+1

|(−2πix)α ϕRh(x)|

≤Cn(1+ |x|)−(n+1) ∑|α |≤n+1

∥∥(∂α(ϕRh))∧∥∥

L∞

≤Cn(1+ |x|)−(n+1) ∑|α |≤n+1

∥∥∂α(ϕRh)∥∥

L1

≤Cn(2nRnvn)

1/q′(1+ |x|)−(n+1) ∑|α |≤n+1

∥∥∂α(ϕRh)∥∥

Lq

≤Cn,R(1+ |x|)−(n+1) ∑|α |≤n+1

∥∥∂αh∥∥

Lq ,

where we used Leibniz’s rule (Proposition 2.3.22 (14)) and the fact that all deriva-

tives of ϕR are pointwise bounded by constants depending on R.

Integrate the previously displayed inequality with respect to x to obtain

∥∥ϕRh∥∥

L1 ≤CR,n ∑|α |≤n+1

∥∥∂αh∥∥

Lq < ∞ . (2.5.9)

Therefore, Fourier inversion holds for ϕRh (see Exercise 2.2.6). This implies that

ϕRh is equal a.e. to a continuous function, namely the inverse Fourier transform of

its Fourier transform. Since ϕR = 1 on the ball B(0,R), we conclude that h is a.e.

equal to a continuous function in this ball. Since R > 0 was arbitrary, it follows that


h is a.e. equal to a continuous function on Rn, which we denote by H. Finally, (2.5.2)

is a direct consequence of (2.5.9) with R = 1, since |H(0)| ≤ ‖ϕ1h‖L1 .

2.5.2 The Transpose and the Adjoint of a Linear Operator

We briefly discuss the notions of the transpose and the adjoint of a linear operator.

We first recall real and complex inner products. For f ,g measurable functions on

Rn, we define the complex inner product

⟨f |g

⟩=

∫

Rnf (x)g(x)dx ,

whenever the integral converges absolutely. We reserve the notation

⟨f ,g

⟩=

∫

Rnf (x)g(x)dx

for the real inner product on L2(Rn) and also for the action of a distribution f on

a test function g. (This notation also makes sense when a distribution f coincides

with a function.)

Let 1≤ p,q≤ ∞. For a bounded linear operator T from Lp(X ,µ) to Lq(Y,ν) we

denote by T ∗ its adjoint operator defined by

⟨T ( f ) |g

⟩=

∫

YT ( f )gdν =

∫

Xf T ∗(g)dµ =

⟨f |T ∗(g)

⟩(2.5.10)

for f in Lp(X ,µ) and g in Lq′(Y,ν) (or in a dense subspace of it). We also define the

transpose of T as the unique operator T t that satisfies

⟨T ( f ),g

⟩=

∫

YT ( f )gdx =

∫

Xf T t(g)dx =

⟨f ,T t(g)

⟩

for all f ∈ Lp(X ,µ) and all g ∈ Lq′(Y,ν).If T is an integral operator of the form

T ( f )(x) =∫

XK(x,y) f (y)dµ(y),

then T ∗ and T t are also integral operators with kernels K∗(x,y) = K(y,x) and

Kt(x,y) = K(y,x), respectively. If T has the form T ( f ) = ( f m)∨ , that is, it is given

by multiplication on the Fourier transform by a (complex-valued) function m(ξ ),

then T ∗ is given by multiplication on the Fourier transform by the function m(ξ ).Indeed for f ,g in S (Rn) we have

∫

Rnf T ∗(g)dx =

∫

RnT ( f ) gdx

=∫

RnT ( f ) gdξ

=∫

Rnf m gdξ

=

∫

Rnf (mg)∨ dx .

A similar argument (using Theorem 2.2.14 (5)) gives that if T is given by multipli-

cation on the Fourier transform by the function m(ξ ), then T t is given by multipli-

cation on the Fourier transform by the function m(−ξ ). Since the complex-valued

functions m(ξ ) and m(−ξ ) may be different, the operators T ∗ and T t may be dif-

ferent in general. Also, if m(ξ ) is real-valued, then T is self-adjoint (i.e., T = T ∗)while if m(ξ ) is even, then T is self-transpose (i.e., T = T t ).

2.5.3 The Spaces M p,q(Rn)

Definition 2.5.5. Given 1≤ p,q≤∞, we denote by M p,q(Rn) the set of all bounded

linear operators from Lp(Rn) to Lq(Rn) that commute with translations.

By Theorem 2.5.2 we have that every T in M p,q is given by convolution with a

tempered distribution. We introduce a norm on M p,q by setting

∥∥T∥∥

M p,q =∥∥T

∥∥Lp→Lq ,

that is, the norm of T in M p,q is the operator norm of T as an operator from Lp to

Lq. It is a known fact that under this norm, M p,q is a complete normed space (i.e.,

a Banach space).

Next we show that when p> q the set M p,q consists of only one element, namely

the zero operator T = 0. This means that the only interesting classes of operators

arise when p≤ q.

Theorem 2.5.6. M p,q = 0 whenever 1≤ q < p < ∞.

Proof. Let f be a nonzero C ∞0 function and let h ∈ Rn. We have

∥∥τh(T ( f ))+T ( f )∥∥

Lq =∥∥T (τh( f )+ f )

∥∥Lq ≤

∥∥T∥∥

Lp→Lq

∥∥τh( f )+ f∥∥

Lp .

Now let |h| → ∞ and use Exercise 2.5.1. We conclude that

21q∥∥T ( f )

∥∥Lq ≤

∥∥T∥∥

Lp→Lq 21p∥∥ f

∥∥Lp ,

which is impossible if q < p unless T is the zero operator.

Next we have a theorem concerning the duals of the spaces M p,q(Rn).

Theorem 2.5.7. Let 1 < p ≤ q < ∞ and T ∈M p,q(Rn). Then T can be defined on

Lq′(Rn), coinciding with its previous definition on the subspace Lp(Rn)∩Lq′(Rn) of

Lp(Rn), so that it maps Lq′(Rn) to Lp′(Rn) with norm

∥∥T∥∥

Lq′→Lp′ =∥∥T

∥∥Lp→Lq . (2.5.11)

In other words, we have the following isometric identification of spaces:

Mq′,p′(Rn) = M

p,q(Rn) .

Proof. We first observe that if T : Lp→ Lq is given by convolution with u∈S ′, then

the adjoint operator T ∗ : Lq′ → Lp′ is given by convolution with u ∈S ′. Indeed, for

f ,g ∈S (Rn) we have

∫

Rnf T ∗(g)dx =

∫

RnT ( f )gdx

=∫

Rn( f ∗u)gdx

=∫

Rnf (g∗ u)dx

=∫

Rnf g∗ udx .

Therefore T ∗ is given by convolution with u when applied to Schwartz functions.

Next we observe the validity of the identity

f ∗ u = ( f ∗u)˜, f ∈S . (2.5.12)

It remains to show that T (convolution with u) and T ∗ (convolution with u ) map Lq′

to Lp′ with the same norm. But this easily follows from (2.5.12), which implies that

∥∥ f ∗ u∥∥

Lp′∥∥ f∥∥

Lq′=

∥∥ f ∗u∥∥

Lp′∥∥ f

∥∥Lq′

,

for all nonzero Schwartz functions f . We conclude that

‖T ∗‖Lq′→Lp′ = ‖T‖Lq′→Lp′

and therefore

‖T‖Lp→Lq = ‖T‖Lq′→Lp′ .

This establishes the claimed assertion.

We next focus attention on the spaces M p,q(Rn) whenever p = q. These spaces

are of particular interest, since they include the singular integral operators, which

we study in Chapter 5.


2.5.4 Characterizations of M 1,1(Rn) and M 2,2(Rn)

It would be desirable to have a characterization of the spaces M p,p in terms of

properties of the convolving distribution. Unfortunately, this is unknown at present

(it is not clear whether it is possible) except for certain cases.

Theorem 2.5.8. An operator T is in M 1,1(Rn) if and only if it is given by convo-

lution with a finite Borel (complex-valued) measure. In this case, the norm of the

operator is equal to the total variation of the measure.

Proof. If T is given with convolution with a finite Borel measure µ , then clearly T

maps L1 to itself and ‖T‖L1→L1 ≤ ‖µ‖M , where ‖µ‖M is the total variation of µ .

Conversely, let T be an operator bounded from L1 to L1 that commutes with trans-

lations. By Theorem 2.5.2, T is given by convolution with a tempered distribution

u. Let

fε(x) = ε−ne−π|x/ε |2

.

Since the functions fε are uniformly bounded in L1, it follows from the boundedness

of T that fε ∗u are also uniformly bounded in L1. Since L1 is naturally embedded in

the space of finite Borel measures, which is the dual of the space C00 of continuous

functions that tend to zero at infinity, we obtain that the family fε ∗u lies in a fixed

multiple of the unit ball of C ∗00. By the Banach–Alaoglu theorem, this is a weak∗

compact set. Therefore, some subsequence of fε ∗u converges in the weak∗ topology

to a measure µ . That is, for some εk → 0 and all g ∈ C00(Rn) we have

limk→∞

∫

Rng(x)( fεk

∗u)(x)dx =∫

Rng(x)dµ(x) . (2.5.13)

We claim that u = µ . To see this, fix g ∈S . Equation (2.5.13) implies that

⟨u, fεk

∗g⟩=

⟨u, fεk

∗g⟩→

⟨µ ,g

⟩

as k→ ∞. Exercise 2.3.2 gives that g∗ fεkconverges to g in S . Therefore,

⟨u, fεk

∗g⟩→

⟨u,g

⟩.

It follows from (2.5.13) that 〈u,g〉= 〈µ ,g〉, and since g was arbitrary, u = µ .

Next, (2.5.13) implies that for all g ∈ C00 we have

∣∣∣∣∫

Rng(x)dµ(x)

∣∣∣∣≤∥∥g

∥∥L∞

supk

∥∥ fεk∗u

∥∥L1 ≤

∥∥g∥∥

L∞

∥∥T∥∥

L1→L1 . (2.5.14)

The Riesz representation theorem gives that the norm of the functional

g →∫

Rng(x)dµ(x)

on C00 is exactly ‖µ‖M . It follows from (2.5.14) that ‖T‖L1→L1 ≥ ‖µ‖M . Since the

reverse inequality is obvious, we conclude that ‖T‖L1→L1 = ‖µ‖M .


Let µ be a finite Borel measure. The operator h → h ∗ µ maps Lp(Rn) to itself

for all 1≤ p≤∞; hence M 1,1(Rn) can be identified with a subspace of M ∞,∞(Rn).But there exist bounded linear operators Φ on L∞ that commute with translations

for which there does not exist a finite Borel measure µ such that Φ(h) = h ∗ µ for

all h ∈ L∞(Rn). The following example captures such a behavior.

Example 2.5.9. Let (X ,‖ · ‖L∞) be the space of all complex-valued bounded func-

tions on the real line such that

Φ( f ) = limR→+∞

1

R

∫ R

0f (t)dt

exists. Then Φ is a bounded linear functional on X with norm 1 and has a bounded

extension Φ on L∞ with norm 1, by the Hahn–Banach theorem. We may view Φas a bounded linear operator from L∞(R) to the space of constant functions, which

is contained in L∞(R). We note that Φ commutes with translations, since for all

f ∈ L∞(R) and x ∈ R we have

Φ(τx( f ))− τx(Φ( f )) = Φ(τx( f ))− Φ( f ) = Φ(τx( f )− f ) =Φ(τx( f )− f ) = 0,

where the last two equalities follow from the fact that for L∞ functions f the expres-

sion 1R

∫ R0 ( f (t−x)− f (t))dt is bounded by

|x|R‖ f‖L∞ when R > |x| and thus tends to

zero as R→ ∞. If Φ(ϕ) = ϕ ∗u for some u ∈S ′(Rn) and all ϕ ∈S (Rn), since Φvanishes on S , the uniqueness in Theorem 2.5.2 yields that u = 0. Hence, if there

existed a finite Borel measure µ such that Φ(h) = h∗µ all h ∈ L∞, in particular we

would have 0 = Φ(ϕ) = ϕ ∗ µ for all ϕ ∈S , hence µ would be the zero measure.

But obviously, this is not the case, since Φ is not the zero operator on X .

We now study the case p = 2. We have the following theorem:

Theorem 2.5.10. An operator T is in M 2,2(Rn) if and only if it is given by convo-

lution with some u ∈S ′ whose Fourier transform u is an L∞ function. In this case

the norm of T : L2 → L2 is equal to ‖u‖L∞ .

Proof. If u ∈ L∞, Plancherel’s theorem gives

∫

Rn| f ∗u|2 dx =

∫

Rn| f (ξ )u(ξ )|2 dξ ≤

∥∥u∥∥2

L∞

∥∥ f∥∥2

L2 ;

therefore, ‖T‖L2→L2 ≤ ‖u‖L∞ , and hence T is in M 2,2(Rn).Now suppose that T ∈M 2,2(Rn) is given by convolution with a tempered distri-

bution u. We show that u is a bounded function. For R > 0 let ϕR be a C ∞0 function

supported inside the ball B(0,2R) and equal to one on the ball B(0,R). The product

of the function ϕR with the distribution u is ϕRu = ((ϕR)∨ ∗ u)= T (ϕ∨R ), which

is an L2 function. Since the L2 function ϕRu coincides with the distribution u on

the set B(0,R), it follows that u is in L2(B(0,R)) for all R > 0 and therefore it is

in L2loc. If f ∈ L∞(Rn) has compact support, the function f u is in L2, and therefore

Plancherel’s theorem and the boundedness of T give

∫

Rn| f (x)u(x)|2 dx =

∫

Rn|T ( f∨)(x)|2 dx≤

∥∥T∥∥2

L2→L2

∫

Rn| f (x)|2 dx .

We conclude that for all bounded functions with compact support f we have

∫

Rn

(‖T‖2

L2→L2 −|u(x)|2)| f (x)|2 dx≥ 0 .

Taking f (x1, . . . ,xn) = (2r)−n/2∏nj=1 χ[−r,r](x j) for r > 0 and using Corollary 2.1.16,

we obtain that ‖T‖2L2→L2−|u(x)|2≥ 0 for almost all x. Hence u is in L∞ and ‖u‖L∞ ≤

‖T‖L2→L2 . Combining this with the estimate ‖T‖L2→L2 ≤ ‖u‖L∞ , which holds if

u ∈ L∞, we deduce that ‖T‖L2→L2 = ‖u‖L∞ .

2.5.5 The Space of Fourier Multipliers Mp(Rn)

We have now characterized all convolution operators that map L2 to L2. Suppose

now that T is in M p,p, where 1 < p < 2. As discussed in Theorem 2.5.7, T also

maps Lp′ to Lp′ . Since p < 2 < p′, by Theorem 1.3.4, it follows that T also maps L2

to L2. Thus T is given by convolution with a tempered distribution whose Fourier

transform is a bounded function.

Definition 2.5.11. Given 1≤ p<∞, we denote by Mp(Rn) the space of all bounded

functions m on Rn such that the operator

Tm( f ) = ( f m)∨, f ∈S ,

is bounded on Lp(Rn) (or is initially defined in a dense subspace of Lp(Rn) and has

a bounded extension on the whole space). The norm of m in Mp(Rn) is defined by

∥∥m∥∥

Mp=

∥∥Tm

∥∥Lp→Lp . (2.5.15)

Definition 2.5.11 implies that m ∈ Mp if and only if Tm ∈ M p,p. Elements of

the space Mp are called Lp multipliers or Lp Fourier multipliers. It follows from

Theorem 2.5.10 that M2, the set of all L2 multipliers, is L∞. Theorem 2.5.8 implies

that M1(Rn) is the set of the Fourier transforms of finite Borel measures that is

usually denoted by M (Rn). Theorem 2.5.7 states that a bounded function m is an

Lp multiplier if and only if it is an Lp′ multiplier, and in this case

∥∥m∥∥

Mp=

∥∥m∥∥

Mp′, 1 < p < ∞ .

It is a consequence of Theorem 1.3.4 that the normed spaces Mp are nested, that is,

for 1≤ p≤ q≤ 2 we have

M1 Mp Mq M2 = L∞.

Moreover, if m ∈Mp and 1≤ p≤ 2≤ p′, Theorem 1.3.4 gives

∥∥Tm

∥∥L2→L2 ≤

∥∥Tm

∥∥ 12

Lp→Lp

∥∥Tm

∥∥ 12

Lp′→Lp′ =∥∥Tm

∥∥Lp→Lp , (2.5.16)

since 1/2 = (1/2)/p+(1/2)/p′. Theorem 1.3.4 also gives that

∥∥m∥∥

Mp≤

∥∥m∥∥

Mq

whenever 1 ≤ q ≤ p ≤ 2. Thus the Mp’s form an increasing family of spaces as p

increases from 1 to 2.

Example 2.5.12. The function m(ξ ) = e2πiξ ·b is an Lp multiplier for all b ∈ Rn,

since the corresponding operator Tm( f )(x)= f (x+b) is bounded on Lp(Rn). Clearly

‖m‖Mp= 1.

Proposition 2.5.13. For 1 ≤ p < ∞, the normed space(Mp,‖ · ‖Mp

)is a Banach

space. Furthermore, Mp is closed under pointwise multiplication and is a Banach

algebra.

Proof. It suffices to consider the case 1≤ p≤ 2. It is straightforward that if m1, m2

are in Mp and b ∈ C then m1 +m2 and bm1 are also in Mp. Observe that m1m2 is

the multiplier that corresponds to the operator Tm1Tm2

= Tm1m2and thus

∥∥m1m2

∥∥Mp

=∥∥Tm1

Tm2

∥∥Lp→Lp ≤

∥∥m1

∥∥Mp

∥∥m2

∥∥Mp

.

This proves that Mp is an algebra. To show that Mp is a complete normed space,

consider a Cauchy sequence m j in Mp. It follows from (2.5.16) that m j is Cauchy in

L∞, and hence it converges to some bounded function m in the L∞ norm; moreover

all the m j are a.e. bounded by some constant C uniformly in j. We have to show that

m ∈Mp. Fix f ∈S . We have

Tm j( f )(x) =

∫

Rnf (ξ )m j(ξ )e

2πix·ξ dξ →∫

Rnf (ξ )m(ξ )e2πix·ξ dξ = Tm( f )(x)

a.e. by the Lebesgue dominated convergence theorem, since C | f | is an integrable

upper bound of all integrands on the left in the preceding expression. Since m j j

is a Cauchy sequence in Mp, it is bounded in Mp, and thus sup j ‖m j‖Mp<+∞. An

application of Fatou’s lemma yields that

∫

Rn|Tm( f )|p dx =

∫

Rnliminf

j→∞|Tm j

( f )|p dx

≤ liminfj→∞

∫

Rn|Tm j

( f )|p dx

≤ liminfj→∞

∥∥m j

∥∥p

Mp

∥∥ f∥∥p

Lp ,

which implies that m ∈ Mp. This argument shows that if m j ∈Mp and m j → m

uniformly, then m is in Mp and satisfies

‖m‖Mp≤ liminf

j→∞‖m j‖Mp

.

Apply this inequality to mk−m j in place of m j and mk−m in place of m, for some

fixed k. We obtain

‖mk−m‖Mp≤ liminf

j→∞‖mk−m j‖Mp

(2.5.17)

for each k. Given ε > 0, by the Cauchy criterion, there is an N such that for j,k > N

we have ‖mk−m j‖Mp< ε . Using (2.5.17) we conclude that ‖mk−m‖Mp

≤ ε when

k > N, thus mk converges to m in Mp.

This proves that Mp is a Banach space.

The following proposition summarizes some simple properties of multipliers.

Proposition 2.5.14. For all m ∈Mp, 1≤ p < ∞, x ∈ Rn, and h > 0 we have

∥∥τx(m)∥∥

Mp=

∥∥m∥∥

Mp, (2.5.18)

∥∥δ h(m)∥∥

Mp=

∥∥m∥∥

Mp, (2.5.19)

∥∥m∥∥

Mp=

∥∥m∥∥

Mp,

∥∥e2πi( ·)·xm∥∥

Mp=

∥∥m∥∥

Mp,

∥∥mA∥∥

Mp=

∥∥m∥∥

Mp, A is an orthogonal matrix.

Proof. See Exercise 2.5.2.

Example 2.5.15. We show that for−∞<a<b<∞we have ‖χ[a,b]‖Mp= ‖χ[0,1]‖Mp

.

Indeed, using (2.5.18) we obtain that ‖χ[a,b]‖Mp= ‖χ[0,b−a]‖Mp

, and the latter is

equal to ‖χ[0,1]‖Mpin view of (2.5.19). The fact that we have ‖χ[0,1]‖Mp

< ∞ for all

1 < p < ∞ is shown in Chapter 5.

We continue with the following interesting result.

Theorem 2.5.16. Suppose that m(ξ ,η) ∈Mp(Rn+m), where 1 < p < ∞. Then for

almost every ξ ∈ Rn the function η → m(ξ ,η) is in Mp(Rm), with

∥∥m(ξ , ·)∥∥

Mp(Rm)≤

∥∥m∥∥

Mp(Rn+m).

Proof. Since m lies in L∞(Rn+m), it follows by Fubini’s theorem that for almost all

ξ ∈ Rn, the function η → m(ξ ,η) lies in L∞(Rm) and

∥∥m(ξ , ·)∥∥

L∞(Rm)≤

∥∥m∥∥

L∞(Rn+m). (2.5.20)


Fix f1, g1 in S (Rn) and f2, g2 in S (Rm). Define the functions ( f1⊗ f2)(x,y) =f1(x) f2(y) when x ∈ Rn and y ∈ Rm. For all ξ for which (2.5.20) is satisfied define

M(ξ ) =∫

Rm

(m(ξ , ·) f2

)∨(y)g2(y)dy =

∫

Rmm(ξ ,η) f2(η)g2

∨(η)dη

and observe that∣∣∣∣∫

Rn

(M( ·) f1

)∨(x)g1(x)dx

∣∣∣∣ =

∣∣∣∣∫

RnM(ξ ) f1(ξ )g1

∨(ξ )dξ

∣∣∣∣

=

∣∣∣∣∫∫

Rn+mm(ξ ,η) f1⊗ f2(ξ ,η)(g1⊗g2)

∨(ξ ,η)dξ dη

∣∣∣∣

=

∣∣∣∣∫∫

Rn+m(m f1⊗ f2)

∨(x,y)(g1⊗g2)(x,y)dxdy

∣∣∣∣≤

∥∥m∥∥

Mp(Rn+m)

∥∥ f1

∥∥Lp

∥∥ f2

∥∥Lp

∥∥g1

∥∥Lp′

∥∥g2

∥∥Lp′ .

In view of the identity

∥∥(M( ·) f1)∨∥∥

Lp = sup‖g1‖

Lp′≤1

∣∣∣∣∫

Rn

(M( ·) f1

)∨(x)g1(x)dx

∣∣∣∣ ,

it follows that, for the ξ that satisfy (2.5.20), M(ξ ) lies in Mp(Rn) with

∥∥M∥∥

Mp(Rn)≤

∥∥m∥∥

Mp(Rn+m)

∥∥ f2

∥∥Lp

∥∥g2

∥∥Lp′ .

Since ‖M‖L∞ ≤ ‖M‖Mpfor almost all ξ ∈ Rn, we obtain

∣∣∣∣∫

Rm(m(ξ , ·) f2)

∨(y)g2(y)dy

∣∣∣∣= |M(ξ )| ≤∥∥m

∥∥Mp(Rn+m)

∥∥ f2

∥∥Lp

∥∥g2

∥∥Lp′ , (2.5.21)

which of course implies the required conclusion, by taking the supremum over all

g2 in Lp′ with norm at most 1.

Example 2.5.17. (The cone multiplier) On Rn+1 define the function

mλ (ξ1, . . . ,ξn+1) =

(1− ξ 2

1 + · · ·+ξ 2n

ξ 2n+1

)λ

+

, λ > 0,

where the plus sign indicates that mλ = 0 if the expression inside the parentheses is

negative. The multiplier mλ is called the cone multiplier with parameter λ . If mλ is

in Mp(Rn+1), then the function bλ (ξ ) = (1−|ξ |2)λ+ defined on Rn is in Mp(R

n).Indeed, by Theorem 2.5.16 we have that for some ξn+1 = h, bλ (ξ1/h, . . . ,ξn/h) is in

Mp(Rn) and hence so is bλ by property (2.5.19).

Exercises

2.5.1. Prove that if f ∈ Lq(Rn) and 0 < q < ∞, then

∥∥τh( f )+ f∥∥

Lq → 21/q∥∥ f

∥∥Lq as |h| → ∞.

2.5.2. Prove Proposition 2.5.14. Also prove that if δh j

j is a dilation operator in the

jth variable (for instance δ h11 f (x) = f (h1x1,x2, . . . ,xn)), then

∥∥δ h11 · · ·δ hn

n m∥∥

Mp=

∥∥m∥∥

Mp.

2.5.3. Let m ∈Mp(Rn) where 1≤ p < ∞.

(a) If ψ is a function on Rn whose inverse Fourier transform is an integrable func-

tion, then prove that ∥∥ψm∥∥

Mp≤

∥∥ψ∨∥∥

L1

∥∥m∥∥

Mp.

(b) If ψ is in L1(Rn), then prove that

∥∥ψ ∗m∥∥

Mp≤

∥∥ψ∥∥

L1

∥∥m∥∥

Mp.

2.5.4. Fix a multi-index γ .

(a) Prove that the map T ( f ) = f ∗∂ γδ0 maps S continuously into S .

(b) Prove that when 1/p−1/q = |γ |/n, T does not extend to an element of the space

M p,q.

2.5.5. Let Kγ(x) = |x|−n+γ , where 0 < γ < n. Use Theorem 1.4.25 to show that the

operator

Tγ( f ) = f ∗Kγ , f ∈S ,

extends to a bounded operator in M p,q(Rn), where 1/p−1/q= γ/n, 1< p< q<∞.

This provides an example of a nontrivial operator in M p,q(Rn) when p < q.

2.5.6. (a) Use the ideas of the proof of Proposition 2.5.13 to show that if m j ∈Mp,

1≤ p < ∞, ‖m j‖Mp≤C for all j = 1,2, . . . , and m j → m a.e., then m ∈Mp and

∥∥m∥∥

Mp(Rn)≤ liminf

j→∞

∥∥m j

∥∥Mp(Rn)

≤C .

(b) Prove that if m ∈Mp, 1≤ p < ∞, and the limit m0(ξ ) = limR→∞

m(ξ/R) exists for

all ξ ∈Rn, then m0 is a radial function in Mp(Rn) and satisfies ‖m0‖Mp

≤ ‖m‖Mp.

(c) If m ∈Mp(R) has left and right limits at the origin, then prove that

∥∥m∥∥

Mp(R)≥max(|m(0+)|, |m(0−)|).

(d) Suppose that for some 1≤ p < ∞, mt ∈Mp(Rn) for all 0 < t < ∞. Prove that

∫ ∞

0

∥∥mt

∥∥Mp(Rn)

dt

t< ∞ =⇒ m(ξ ) =

∫ ∞

0mt(ξ )

dt

t∈Mp .

2.5.7. Let 1≤ p <∞ and suppose that m∈Mp(Rn) satisfies |m(ξ )| ≥ c(1+ |ξ |)−N

for some c,N > 0. Prove that the operator T ( f ) = ( f m−1)∨ satisfies ‖T ( f )‖Lp ≥cp‖ f‖Lp for all f ∈S (Rn), where cp = ‖m‖−1

Mp.

2.5.8. (a) Prove that if m ∈ L∞(Rn) satisfies m∨ ≥ 0, then for all 1≤ p <∞ we have

∥∥m∥∥

Mp=

∥∥m∨∥∥

L1 .

(b) (L. Colzani and E. Laeng) On the real line let

m1(ξ ) =

−1 for ξ > 0

1 for ξ < 0,m2(ξ ) =

min(ξ −1,0) for ξ > 0

max(ξ +1,0) for ξ < 0.

Prove that ∥∥m1

∥∥Mp

=∥∥m2

∥∥Mp

for all 1 < p < ∞.[Hint: Part (a): Use Exercise 1.2.9. Part (b): Use part (a) to show that ‖m2m−1

1 ‖Mp=

1. Deduce that ‖m2‖Mp≤ ‖m1‖Mp

. For the converse use Exercise 2.5.6 (c).]

2.5.9. ([94]) Let 1 < p <∞ and 0 < A <∞. Prove that the following are equivalent:

(a) The operator f → ∑m∈Zn am f (x−m) is bounded on Lp(Rn) with norm A.

(b) The Mp norm of the function ∑m∈Zn ame−2πim·x is exactly A.

(c) The operator given by convolution with the sequence am is bounded on ℓp(Zn)with norm A.

2.5.10. ([177]) Let m(ξ ) in Mp(Rn) be supported in [0,1]n. Then the periodic ex-

tension of m in Rn,

M(ξ ) = ∑k∈Zn

m(ξ − k) ,

is also in Mp(Rn).

2.5.11. Suppose that u is a C ∞ function on Rn \0 that is homogeneous of degree

−n+ iτ , τ ∈R. Prove that the operator given by convolution with u maps L2(Rn) to

L2(Rn).

2.5.12. ([142]) Let m1 ∈ Lr(Rn) and m2 ∈ Lr′(Rn) for some 2 ≤ r ≤ ∞. Prove that

m1 ∗m2 ∈Mp(Rn) when 1

p− 1

2= 1

rand 1≤ p≤ 2.[

Hint: Prove that the trilinear operator (m1,m2, f ) →((m1∗m2) f

)∨ is bounded from

L2×L2×L1 → L1 and L∞×L1×L2 → L2. Apply trilinear complex interpolation

(Corollary 7.2.11 in [131]) to deduce the required conclusion for 1≤ p≤ 2.]

2.6 Oscillatory Integrals 161

2.5.13. Show that the function ei|ξ |2 is an Lp Fourier multiplier on Rn if and only if

p = 2.[Hint: By Exercise 2.4.12 the inverse Fourier transform of ei|ξ |2 is in L∞, thus the

operator f →(

f (ξ )eiπ|ξ |2)∨ maps L1 to L∞. Since this operator also maps L2 to L2,

it should map Lp to Lp′ for all 1≤ p≤ 2.]

2.6 Oscillatory Integrals

Oscillatory integrals have played an important role in harmonic analysis from its

outset. The Fourier transform is the prototype of oscillatory integrals and provides

the simplest example of a nontrivial phase, a linear function of the variable of in-

tegration. More complicated phases naturally appear in the subject; for instance,

Bessel functions provide examples of oscillatory integrals in which the phase is a

sinusoidal function.

In this section we take a quick look at oscillatory integrals. We mostly concentrate

on one-dimensional results, which already require some significant analysis. We

examine only a very simple higher-dimensional situation. Our analysis here is far

from adequate.

Definition 2.6.1. An oscillatory integral is an expression of the form

I(λ ) =∫

Rneiλϕ(x)ψ(x)dx , (2.6.1)

where λ is a positive real number, ϕ is a real-valued function on Rn called the

phase, and ψ is a complex-valued and smooth integrable function on Rn, which is

often taken to have compact support.

2.6.1 Phases with No Critical Points

We begin by studying the simplest possible one-dimensional case. Suppose that ϕand ψ are smooth functions on the real line such that suppψ is a closed interval and

ϕ ′(x) = 0 for all x ∈ suppψ .

Since ϕ ′ has no zeros, it must be either strictly positive or strictly negative every-

where on the support of ψ . It follows that ϕ is monotonic on the support of ψ and

we are allowed to change variables

u = ϕ(x)


in (2.6.1). Then dx= (ϕ ′(x))−1du= (ϕ−1)′(u)du, where ϕ−1 is the inverse function

of ϕ . We transform the integral in (2.6.1) into

∫

Reiλuψ(ϕ−1(u))(ϕ−1)′(u)du (2.6.2)

and we note that the function θ(u) = ψ(ϕ−1(u))(ϕ−1)′(u) is smooth and has com-

pact support on R. We therefore interpret the integral in (2.6.1) as θ(−λ/2π), where

θ is the Fourier transform of θ . Since θ is a smooth function with compact support,

it follows that the integral in (2.6.2) has rapid decay as λ → ∞.

A quick way to see that the expression θ(−λ/2π) has decay of order λ−N for all

N > 0 as λ tends to ∞ is the following. Write

eiλu =1

(iλ )N

dN

duN(eiλu)

and integrate by parts N times to express the integral in (2.6.2) as

(−1)N

(iλ )N

∫

Reiλu dNθ(u)

duNdu ,

from which the assertion follows. Hence

|I(λ )|= |θ(−λ/2π)| ≤CNλ−N , (2.6.3)

where CN = ‖θ (N)‖L1 , which depends on derivatives of ϕ and ψ .

We now turn to a higher-dimensional analogue of this situation.

Definition 2.6.2. We say that a point x0 is a critical point of a phase function ϕ if

∇ϕ(x0) =(∂1ϕ(x0), . . . ,∂nϕ(x0)

)= 0 .

Example 2.6.3. Let ξ ∈ Rn \ 0. Then the phase functions ϕ1(x) = x · ξ , ϕ2(x) =ex·ξ have no critical points, while the phase function ϕ3(x) = |x|2− x · ξ has one

critical point at x0 =12ξ .

The next result concerns the behavior of oscillatory integrals whose phase func-

tions have no critical points.

Proposition 2.6.4. Suppose that ψ is a compactly supported smooth function on Rn

and that ϕ is a real-valued C ∞ function on Rn that has no critical points on the

support of ψ . Then the oscillatory integral

I(λ ) =∫

Rneiλϕ(x)ψ(x)dx (2.6.4)

obeys a bound of the form |I(λ )| ≤CNλ−N for all λ ≥ 1 and all N > 0, where CN

depends on N and on ϕ and ψ .


Proof. Since the case n = 1 has already been discussed, we concentrate on dimen-

sions n≥ 2. For each y in the support of ψ there is a unit vector θy such that

θy ·∇ϕ(y) = |∇ϕ(y)| .

By the continuity of ∇ϕ there is a small neighborhood B(y,ry) of y such that for all

x ∈ B(y,ry) we have

θy ·∇ϕ(x)≥1

2|∇ϕ(y)|> 0 .

Cover the support of ψ by a finite number of balls B(y j,ry j), j = 1, . . . ,m, and pick

c = min j12|∇ϕ(y j)|; we have

θy j·∇ϕ(x)≥ c > 0 (2.6.5)

for all x ∈ B(y j,ry j) and j = 1, . . . ,m.

Next we find a smooth partition of unity of Rn such that each member ζk of the

partition is supported in some ball B(y j,ry j) or lies outside the support of ψ . We

therefore write

I(λ ) =∑k

∫

Rneiλϕ(x)ψ(x)ζk(x)dx , (2.6.6)

where the sum contains only a finite number of indices, since only a finite number

of the ζk’s meet the support of ψ . It suffices to show that every term in the sum in

(2.6.6) has rapid decay in λ as λ → ∞.

To this end, we fix a k and we pick a j such that the support of ψζk is contained

in some ball B(y j,ry j). We find unit vectors θy j ,2, . . . ,θy j ,n, such that the system

θy j,θy j ,2, . . . ,θy j ,n is an orthonormal basis of Rn. Let e j be the unit (column)

vector on Rn whose jth coordinate is one and whose remaining coordinates are zero.

We find an orthogonal matrix R such that Rte1 = θy jand we introduce the change

of variables u = y j +R(x− y j) in the integral

Ik(λ ) =∫

Rneiλϕ(x)ψ(x)ζk(x)dx .

The map x → u = (u1, . . . ,un) is a rotation that fixes y j and preserves the ball

B(y j,ry j). Defining ϕ(x) = ϕo(u), ψ(x) = ψo(u), ζk(x) = ζ o

k (u), under this new

coordinate system we write

Ik(λ ) =∫

K

∫

Reiλϕo(u)ψo(u1, . . . ,un)ζ

ok (u1, . . . ,un)du1

du2 · · ·dun , (2.6.7)

where K is a compact subset of Rn−1. Since R is an orthogonal matrix, R−1 = Rt ,

and the change of variables x = y j +Rt(u− y j) implies that

∂x

∂u1= first column of Rt = first row of R = Rte1 = θy j

.


Thus for all x ∈ B(y j,r j) we have

∂ϕo(u)

∂u1=∂ϕ(y j +Rt(u− y j))

∂u1= ∇ϕ(x) · ∂x

∂u1= ∇ϕ(x) ·θy j

≥ c > 0

in view of condition (2.6.5). This lower estimate is valid for all u ∈ B(y j,ry j), and

therefore the inner integral inside the curly brackets in (2.6.7) is at most CNλ−N by

estimate (2.6.3). Integrating over K results in the same conclusion for I(λ ) defined

in (2.6.4).

2.6.2 Sublevel Set Estimates and the Van der Corput Lemma

We discuss a sharp decay estimate for one-dimensional oscillatory integrals. This

estimate is obtained as a consequence of delicate size estimates for the Lebesgue

measures of the sublevel sets |u| ≤α for a function u. In what follows, u(k) denotes

the kth derivative of a function u(t) defined on R, and C k the space of all functions

whose kth derivative exists and is continuous.

Lemma 2.6.5. Let k ≥ 1 and suppose that a0, . . . ,ak are distinct real numbers. Let

a = min(a j) and b = max(a j) and let f be a real-valued C k−1 function on [a,b] that

is C k on (a,b). Then there exists a point y in (a,b) such that

k

∑m=0

cm f (am) = f (k)(y) ,

where cm = (−1)k k!k

∏ℓ=0ℓ =m

(aℓ−am)−1.

Proof. Suppose we could find a polynomial pk(x)=∑kj=0 b jx

j such that the function

ϕ(x) = f (x)− pk(x) (2.6.8)

satisfies ϕ(am) = 0 for all 0 ≤ m ≤ k. Since the a j are distinct, we apply Rolle’s

theorem k times to find a point y in (a,b) such that

f (k)(y) = k!bk .

The existence of a polynomial pk such that (2.6.8) is satisfied is equivalent to the

existence of a solution to the matrix equation

⎛⎜⎜⎜⎜⎜⎝

ak0 ak−1

0 . . . a0 1

ak1 ak−1

1 . . . a1 1...

......

......

akk−1 ak−1

k−1 . . . ak−1 1

akk ak−1

k . . . ak 1

⎞⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎝

bk

bk−1

...

b1

b0

⎞⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎜⎜⎜⎝

f (a0)f (a1)

...

f (ak−1)f (ak)

⎞⎟⎟⎟⎟⎟⎠

.

The determinant of the square matrix on the left is called the Vandermonde deter-

minant and is equal tok−1

∏ℓ=0

k

∏j=ℓ+1

(aℓ−a j) = 0 .

Since the a j are distinct, it follows that the system has a unique solution. Using

Cramer’s rule, we solve this system to obtain

bk =k

∑m=0

(−1)m f (am)

k−1

∏ℓ=0ℓ =m

k

∏j=ℓ+1j =m

(aℓ−a j)

k−1

∏ℓ=0

k

∏j=ℓ+1

(aℓ−a j)

=k

∑m=0

(−1)m f (am)k

∏ℓ=0ℓ =m

(aℓ−am)−1(−1)k−m .

The required conclusion now follows with cm as claimed.

Lemma 2.6.6. Let E be a measurable subset of R with finite nonzero Lebesgue mea-

sure and let k ∈ Z+. Then there exist a0, . . . ,ak in E such that for all ℓ = 0,1, . . . ,kwe have

k

∏j=0j =ℓ

|a j−aℓ| ≥ (|E|/2e)k . (2.6.9)

Proof. Given a measurable set E with finite measure, pick a compact subset E ′ of

E such that |E \E ′| < δ , for some δ > 0. For x ∈ R define T (x) = |(−∞,x)∩E ′|.Then T enjoys the distance-decreasing property

|T (x)−T (y)| ≤ |x− y|

for all x,y ∈ E ′; consequently, by the intermediate value theorem, T is a surjective

map from E ′ to [0, |E ′|]. Let a j be points in E ′ such that T (a j) =jk|E ′| for j =

0,1, . . . ,k. For k an even integer, we have

k

∏j=0j =ℓ

|a j−aℓ| ≥k

∏j=0j =ℓ

∣∣∣ j

k|E ′|− ℓ

k|E ′|

∣∣∣≥k

∏j=0

j = k2

∣∣∣ j

k− 1

2

∣∣∣ |E ′|k =k2−1

∏r=0

( r− k2

k

)2

|E ′|k ,

and it is easily shown that((k/2)!

)2k−k ≥ (2e)−k.

For k an odd integer we have

k

∏j=0j =ℓ

|a j−aℓ| ≥k

∏j=0j =ℓ

∣∣∣ j

k|E ′|− ℓ

k|E ′|

∣∣∣≥k

∏j=0

j = k+12

∣∣∣ j

k− k+1

2k

∣∣∣ |E ′|k ,

while the last product is at least

1

k· 2

k· · ·

k−12

k

2 k+1

2k≥ (2e)−k .

We have therefore proved (2.6.9) with E ′ replacing E. Since |E \E ′|< δ and δ > 0

is arbitrarily small, the required conclusion follows.

The following is the main result of this section.

Proposition 2.6.7. (a) Let u be a real-valued C k function, k ∈ Z+, that satisfies

u(k)(t)≥ 1 for all t ∈ R. Then the following estimate is valid for all α > 0:

∣∣t ∈ R : |u(t)| ≤ α∣∣≤ (2e)((k+1)!)

1k α

1k . (2.6.10)

(b) Let −∞ < a < b < ∞. For all k ≥ 2, for every real-valued C k function u on the

line that satisfies u(k)(t)≥ 1 for all t ∈ [a,b], and every λ ∈ R\0 we have:

∣∣∣∣∫ b

aeiλu(t)dt

∣∣∣∣≤ 12k |λ |− 1k . (2.6.11)

(c) If k = 1, u′(t) is monotonic on (a,b), and u′(t)≥ 1 for all t ∈ (a,b), then for all

nonzero real numbers λ we have

∣∣∣∣∫ b

aeiλu(t)dt

∣∣∣∣≤ 3 |λ |−1 . (2.6.12)

Proof. Part (a): Let E = t ∈ R : |u(t)| ≤ α. If |E| is nonzero, then by Lemma

2.6.6 there exist a0,a1, . . . ,ak in E such that for all ℓ we have

|E|k ≤ (2e)kk

∏j=0j =ℓ

|a j−aℓ| . (2.6.13)

Lemma 2.6.5 implies that there exists y ∈(

mina j,maxa j

)such that

u(k)(y) = (−1)k k!k

∑m=0

u(am)k

∏ℓ=0ℓ =m

(aℓ−am)−1 . (2.6.14)


Using (2.6.13), we obtain that the expression on the right in (2.6.14) is in absolute

value at most

(k+1)! max0≤ j≤k

|u(a j)|(2e)k |E|−k ≤ (k+1)!α (2e)k |E|−k ,

since a j ∈ E. The bound u(k)(t)≥ 1 now implies

|E|k ≤ (k+1)!(2e)kα

as claimed. This proves (2.6.10).

Part (b): We now take k ≥ 2 and we split the interval (a,b) in (2.6.11) into the

sets

R1 = t ∈ (a,b) : |u′(t)| ≤ β ,R2 = t ∈ (a,b) : |u′(t)|> β ,

for some parameter β to be chosen momentarily. The function v = u′ satisfies

v(k−1) ≥ 1 and k−1≥ 1. It follows from part (a) that

∣∣∣∣∫

R1

eiλu(t) dt

∣∣∣∣≤ |R1| ≤ 2e(k!)1

k−1 β1

k−1 ≤ 6kβ1

k−1 .

To obtain the corresponding estimate over R2, we note that if u(k) ≥ 1, then the set

|u′|> β is the union of at most 2k−2 intervals on each of which u′ is monotone.

Let (c,d) be one of these intervals on which u′ is monotone. Then u′ has a fixed sign

on (c,d) and we have

∣∣∣∣∫ d

ceiλu(t) dt

∣∣∣∣ =

∣∣∣∣∫ d

c

(eiλu(t)

)′ 1

λu′(t)dt

∣∣∣∣

≤∣∣∣∣∫ d

ceiλu(t)

( 1

λu′(t)

)′dt

∣∣∣∣+1

|λ |∣∣∣eiλu(d)

u′(d)− eiλu(c)

u′(c)

∣∣∣

≤ 1

|λ |

∫ d

c

∣∣∣( 1

u′(t)

)′∣∣∣dt +2

|λ |β

=1

|λ |

∣∣∣∣∫ d

c

( 1

u′(t)

)′dt

∣∣∣∣+2

|λ |β

=1

|λ |∣∣∣ 1

u′(d)− 1

u′(c)

∣∣∣+ 2

|λ |β

≤ 3

|λ |β ,

where we use the monotonicity of 1/u′(t) in moving the absolute value from inside

the integral to outside. It follows that

∣∣∣∣∫

R2

eiλu(t) dt

∣∣∣∣≤6k

|λ |β .


Choosing β = |λ |−(k−1)/k to optimize and adding the corresponding estimates for

R1 and R2, we deduce the claimed estimate (2.6.11).

Part (c): Repeat the argument in part (b) setting β = 1 and replacing the interval

(c,d) by (a,b).

Corollary 2.6.8. Let (a,b), u(t), λ > 0, and k be as in Proposition 2.6.7. Then for

any function ψ on (a,b) with an integrable derivative and k ≥ 2, we have

∣∣∣∣∫ b

aeiλu(t)ψ(t)dt

∣∣∣∣≤ 12kλ−1/k

[|ψ(b)|+

∫ b

a|ψ ′(s)|ds

].

We also have

∣∣∣∣∫ b

aeiλu(t)ψ(t)dt

∣∣∣∣≤ 3λ−1

[|ψ(b)|+

∫ b

a|ψ ′(s)|ds

],

when k = 1 and u′ is monotonic on (a,b).

Proof. Set

F(x) =∫ x

aeiλu(t) dt

and use integration by parts to write

∫ b

aeiλu(t)ψ(t)dt = F(b)ψ(b)−

∫ b

aF(t)ψ ′(t)dt .

The conclusion easily follows.

Example 2.6.9. The Bessel function of order m is defined as

Jm(r) =1

2π

∫ 2π

0eir sinθ e−imθ dθ .

Here we take both r and m to be real numbers, and we suppose that m>− 12; we refer

to Appendix B for an introduction to Bessel functions and their basic properties.

We use Corollary 2.6.8 to calculate the decay of the Bessel function Jm(r) as

r→ ∞. Set

ϕ(θ) = sin(θ)

and note that ϕ ′(θ) vanishes only at θ = π/2 and 3π/2 inside the interval [0,2π] and

that ϕ ′′(π/2) = −1, while ϕ ′′(3π/2) = 1. We now write 1 = ψ1 +ψ2 +ψ3, where

ψ1 is smooth and compactly supported in a small neighborhood of π/2, and ψ2 is

smooth and compactly supported in a small neighborhood of 3π/2. For j = 1,2,

Corollary 2.6.8 yields

∣∣∣∣∫ 2π

0eir sin(θ)

(ψ j(θ)e

−imθ)dθ

∣∣∣∣≤C mr−1/2


for some constant C, while the corresponding integral containing ψ3 has arbitrary

decay in r in view of estimate (2.6.3) (or Proposition 2.6.4 when n = 1).

Exercises

2.6.1. Suppose that u is a real-valued C k function defined on the line that satisfies

|u(k)(t)| ≥ c0 > 0 for some k ≥ 2 and all t ∈ (a,b). Prove that for λ ∈ R \ 0 we

have ∣∣∣∣∫ b

aeiλu(t) dt

∣∣∣∣≤ 12k (λc0)−1/k

and that the same conclusion is valid when k = 1, provided u′ is monotonic.

2.6.2. Show that if u′ is not monotonic in part (c) of Proposition 2.6.7, then the

conclusion may fail.[Hint: Let ϕ(t) be a real-valued smooth function that is equal to 2t on intervals

[2πk+εk,2π(k+12)−εk] and equal to t on intervals [2π(k+ 1

2)+εk,2π(k+1)−εk],

where 0≤ k ≤ N, for some N ∈ Z+. Show that the absolute value of the integral of

eiϕ(t) over the interval [ε0,2π(N +1)− εN ] tends to infinity as N → ∞.]

2.6.3. Prove that the dependence on k of the constant in part (b) of Proposition 2.6.7

is indeed linear.[Hint: Take u(t) = tk/k! over the interval (0,k!).

]

2.6.4. Follow the steps below to give an alternative proof of part (b) of Proposition

2.6.7. Assume that the statement is known for some k ≥ 2 and some constant C(k)for all intervals [a,b] and all C k functions satisfying u(k) ≥ 1 on [a,b]. Fix a C k+1

function u such that u(k+1) ≥ 1 on an interval [a,b]. Let c be the unique point at

which the function u(k) attains its minimum in [a,b].(a) If u(k)(c) = 0, then for all δ > 0 we have u(k)(t) ≥ δ in the complement of the

interval (c−δ ,c+δ ) and derive the bound

∣∣∣∣∫ b

aeiλu(t)dt

∣∣∣∣≤ 2C(k)(λδ )−1/k +2δ .

(b) If u(k)(c) = 0, then we must have c ∈ a,b. Obtain the bound

∣∣∣∣∫ b

aeiλu(t)dt

∣∣∣∣≤C(k)(λδ )−1/k +δ .

(c) Choose a suitable δ to optimize and deduce the validity of the statement for k+1

with C(k+1) = 2C(k)+2, hence C(k) = 3 ·2k−1 +2k−2, since C(1) = 3.

2.6.5. (a) Prove that for some constant C and all λ ∈ R and ε ∈ (0,1) we have

∣∣∣∣∫

ε≤|t|≤1eiλ t dt

t

∣∣∣∣≤C .

(b) Prove that for some C′ < ∞ , all λ ∈ R, k > 0, and ε ∈ (0,1) we have

∣∣∣∣∫

ε≤|t|≤1eiλ t±tk dt

t

∣∣∣∣≤C′ .

(c) Show that there is a constant C′′ such that for any 0 < ε < N < ∞, for all ξ1,ξ2

in R, and for all integers k ≥ 2, we have

∣∣∣∣∫

ε≤|s|≤Nei(ξ1s+ξ2sk) ds

s

∣∣∣∣≤C′′ .

[Hint: Part (a): For |λ | small use the inequality |eiλ t − 1| ≤ |λ t|. If |λ | is large,

split the domains of integration into the regions |t| ≤ |λ |−1 and |t| ≥ |λ |−1 and use

integration by parts in the second case. Part (b): Write

ei(λ t±tk)−1

t= eiλ t e±itk −1

t+

eiλ t

t

and use part (a). Part (c): When ξ1 = ξ2 = 0 it is trivial. If ξ2 = 0, ξ1 = 0, change

variables t = ξ1s and then split the domain of integration into the sets |t| ≤ 1 and

|t| ≥ 1. In the interval over the set |t| ≤ 1 apply part (b) and over the set |t| ≥ 1 use

integration by parts. In the case ξ2 = 0, change variables t = |ξ2|1/ks and split the

domain of integration into the sets |t| ≥ 1 and |t| ≤ 1. When |t| ≤ 1 use part (b) and

in the case |t| ≥ 1 use Corollary 2.6.8, noting thatdk(ξ1|ξ2|−1/kt±tk)

dt= k!≥ 1.

]

2.6.6. (a) Show that for all a > 0 and λ > 0 the following is valid:

∣∣∣∣∫ aλ

0eiλ log t dt

∣∣∣∣≤ a .

(b) Prove that there is a constant c > 0 such that for all b > λ > 10 we have

∣∣∣∣∫ b

0eiλ t log t dt

∣∣∣∣≤c

λ logλ.

[Hint: Part (b): Consider the intervals (0,δ ) and [δ ,b) for some δ . Apply Proposi-

tion 2.6.7 with k = 1 on one of these intervals and with k = 2 on the other. Then

choose a suitable δ .]

2.6.7. Show that there is a constant C < ∞ such that for all nonintegers γ > 1 and

all λ ,b > 1 we have ∣∣∣∣∫ b

0eiλ tγdt

∣∣∣∣≤C

λ γ.

[Hint: On the interval (0,δ ) apply Proposition 2.6.7 with k = [γ ] + 1 and on the

interval (δ ,b) with k = [γ ]. Then optimize by choosing δ = λ−1/γ .]

HISTORICAL NOTES

The one-dimensional maximal function originated in the work of Hardy and Littlewood [146].

Its n-dimensional analogue was introduced by Wiener [375], who used Lemma 2.1.5, a variant of

the Vitali covering lemma, to derive its Lp boundedness. One may consult the books of de Guzman

[92], [93] for extensions and other variants of such covering lemmas. The actual covering lemma

proved by Vitali [368] says that if a family of closed cubes in Rn has the property that for every point

x ∈ A⊆ Rn there exists a sequence of cubes in the family that tends to x, then it is always possible

to extract a sequence of pairwise disjoint cubes E j from the family such that |A \⋃ j E j| = 0. We

refer to Saks [310] for details and extensions of this theorem.

The class L logL was introduced by Zygmund to give a sufficient condition on the local integra-

bility of the Hardy–Littlewood maximal operator. The necessity of this condition was observed by

Stein [336]. Stein [341] also showed that the Lp(Rn) norm of the centered Hardy–Littlewood max-

imal operator M is bounded above by some dimension-free constant; see also Stein and Stromberg

[345]. Analogous results for maximal operators associated with convex bodies are contained in

Bourgain [35], Carbery [51], and Muller [263]. Bourgain [37] showed the the Hardy-Littlewood

maximal operator associated with cubes is bounded on Lp(Rn) with dimension-free bounds when

p > 1. Aldaz [2] studied the corresponding weak type (1,1) bounds and proved that they grow

to infinity with the dimension; the constant was improved by Aubrun [15]. The situation for the

uncentered maximal operator M on Lp is different, since given any 1 1

such that ‖M‖Lp(Rn)→Lp(Rn) ≥Cnp (see Exercise 2.1.8 for a value of such a constant Cp and also the

article of Grafakos and Montgomery-Smith [136] for a larger value).

The centered maximal function Mµ with respect to a general inner regular locally finite posi-

tive measure µ on Rn is bounded on Lp(Rn,µ) without the additional hypothesis that the measure

is doubling; see Fefferman [117]. The proof of this result requires the following covering lemma,

obtained by Besicovitch [27]: Given any family of closed balls whose centers form a bounded sub-

set of Rn, there exists an at most countable subfamily of balls that covers the set of centers and has

bounded overlap, i.e., no point in Rn belongs to more than a finite number (depending on the dimen-

sion) of the balls in the subfamily. A similar version of this lemma was obtained independently by

Morse [258]. See also Ziemer [385] for an alternative formulation. The uncentered maximal opera-

tor Mµ of Exercise 2.1.1 may not be weak type (1,1) if the measure µ is nondoubling, as shown by

Sjogren [323]; related positive weak type (1,1) results are contained in the article of Vargas [365].

The precise value of the operator norm of the uncentered Hardy–Littlewood maximal function on

Lp(R) was shown by Grafakos and Montgomery-Smith [136] to be the unique positive solution of

the equation (p− 1)xp− pxp−1− 1 = 0. This constant raised to the power n is the operator norm

of the strong maximal function Ms on Lp(Rn) for 1 < p ≤ ∞. The best weak type (1,1) constant

for the centered Hardy–Littlewood maximal operator was shown by Melas [248] to be the largest

root of the quadratic equation 12x2− 22x+ 5 = 0. The strong maximal operator Ms is not weak

type (1,1), but it satisfies the substitute inequality dMs( f )(α) ≤C∫

Rn| f (x)|α (1+ log+

| f (x)|α )n−1 dx.

This result is due to Jessen, Marcinkiewicz, and Zygmund [176], but a geometric proof of it was

obtained by Cordoba and Fefferman [73].

The basic facts about the Fourier transform go back to Fourier [119]. The theory of distributions

was developed by Schwartz [314], [315]. For a concise introduction to the theory of distributions

we refer to Hormander [160] and Yosida [382]. Homogeneous distributions were considered by

Riesz [295] in the study of the Cauchy problem in partial differential equations, although some

earlier accounts are found in the work of Hadamard. They were later systematically studied by

Gelfand and Silov [126], [127]. References on the uncertainty principle include the articles of

Fefferman [114] and Folland and Sitaram [118]. The best possible constant Bp in the Hausdorff–

Young inequality ‖ f ‖Lp′ (Rn)

≤ Bp‖ f‖Lp(Rn) when 1 ≤ p ≤ 2 was shown by Beckner [21] to be

Bp = (p1/p(p′)−1/p′)n/2. This best constant was previously obtained by Babenko [16] in the case

when p′ is an even integer.

A nice treatise of the spaces M p,q is found in Hormander [159]. This reference also con-

tains Theorem 2.5.6, which is due to him. Theorem 2.5.16 is due to de Leeuw [94], but the proof

presented here is taken from Jodeit [178]. De Leeuw’s result in Exercise 2.5.9 says that periodic


elements of Mp(Rn) can be isometrically identified with elements of M (Tn), the latter being the

space of all multipliers on ℓp(Zn). The hint in Exercise 2.5.13 was suggested by M. Peloso.

Parts (b) and (c) of Proposition 2.6.7 are due to van der Corput [364] and are referred to in

the literature as van der Corput’s lemma. The refinement in part (a) was subsequently obtained by

Arhipov, Karachuba, and Cubarikov [8]. The treatment of these results in the text is based on the

article of Carbery, Christ, and Wright [53], which also investigates higher-dimensional analogues

of the theory. Precise asymptotics can be obtained for a variety of oscillatory integrals via the

method of stationary phase; see Hormander [160]. References on oscillatory integrals include the

books of Titchmarsh [362], Erdelyi [107], Zygmund [388], [389], Stein [344], and Sogge [328].

The latter provides a treatment of Fourier integral operators.

Chapter 3

Fourier Series

Principles of Fourier series go back to ancient times. The attempts of the Pythagorean

school to explain musical harmony in terms of whole numbers embrace early ele-

ments of a trigonometric nature. The theory of epicycles in the Almagest of Ptolemy,

based on work related to the circles of Appolonius, contains ideas of astronomical

periodicities that we would interpret today as harmonic analysis. Early studies of

acoustical and optical phenomena, as well as periodic astronomical and geophysical

occurrences, provided a stimulus in the physical sciences toward the rigorous study

of expansions of periodic functions. This study is carefully pursued in this chapter.

The modern theory of Fourier series begins with attempts to solve boundary

value problems using trigonometric functions. The work of d’Alembert, Bernoulli,

Euler, and Clairaut on the vibrating string led to the belief that it might be possible

to represent arbitrary periodic functions as sums of sines and cosines. Fourier an-

nounced belief in this possibility in his solution of the problem of heat distribution

in spatial bodies (in particular, for the cube T3) by expanding an arbitrary function

of three variables as a triple sine series. Fourier’s approach, although heuristic, was

appealing and eventually attracted attention. It was carefully studied and further de-

veloped by many scientists, but most notably by Laplace and Dirichlet, who were

the first to investigate the validity of the representation of a function in terms of its

Fourier series. This is the main topic of study in this chapter.

3.1 Fourier Coefficients

We discuss some basic facts of Fourier analysis on the torus Tn. Throughout this

chapter, n denotes the dimension, i.e., a fixed positive integer.



173

174 3 Fourier Series

3.1.1 The n-Torus Tn

The n-torus Tn is the cube [0,1]n with opposite sides identified. This means that

the points (x1, . . . ,0, . . . ,xn) and (x1, . . . ,1, . . . ,xn) are identified whenever 0 and 1

appear in the same coordinate. A more precise definition can be given as follows:

We say that x,y in Rn are equivalent and we write

x≡ y (3.1.1)

if x− y ∈ Zn. Here Zn is the additive subgroup of all points in Rn with integer

coordinates. It is a simple fact that ≡ is an equivalence relation that partitions Rn

into equivalence classes. The n-torus Tn is then defined as the set Rn/Zn of all such

equivalence classes. When n = 1, this set can be geometrically viewed as a circle

by bending the line segment [0,1] so that its endpoints are brought together. When

n = 2, the identification brings together the left and right sides of the unit square

[0,1]2 as well as the top and bottom sides. The resulting figure is a two-dimensional

manifold embedded in R3 that looks like a donut. See Figure 3.1.

Fig. 3.1 The graph of the

two-dimensional torus T2.

x1

x3

x2

The n-torus is an additive group. The identity element of the group is 0, which

of course coincides with every e j = (0, . . . ,0,1,0, . . . ,0). To avoid multiple appear-

ances of the identity element in the group, we often think of the n-torus as the

set [−1/2,1/2]n. Since the group Tn is additive, the inverse of an element x ∈ Tn

is denoted by −x. For example, −(1/3,1/4) ≡ (2/3,3/4) on T2, or, equivalently,

−(1/3,1/4)− (2/3,3/4) ∈ Z2.

The n-torus Tn can also be thought of as the following subset of Cn,

(e2πix1 , . . . ,e2πixn) ∈ Cn : (x1, . . . ,xn) ∈ [0,1]n , (3.1.2)

in a way analogous to which the unit interval [0,1] can be thought of as the unit

circle in C once 1 and 0 are identified.

Functions on Tn are functions f on Rn that satisfy f (x+m) = f (x) for all x ∈Rn

and m ∈ Zn. Such functions are called 1-periodic in every coordinate. Haar mea-

sure on the n-torus is the restriction of n-dimensional Lebesgue measure to the set

3.1 Fourier Coefficients 175

Tn = [0,1]n. This measure is still denoted by dx, while the measure of a set A⊆ Tn

is denoted by |A|. Translation invariance of Lebesgue measure and the periodicity

of functions on Tn imply that for all integrable functions f on Tn, we have

∫

Tnf (x)dx =

∫

[−1/2,1/2]nf (x)dx =

∫

[a1,1+a1]×···×[an,1+an]f (x)dx (3.1.3)

for any real numbers a1, . . . ,an. In view of periodicity, integration by parts on the

torus does not produce boundary terms; given f ,g continuously differentiable func-

tions on Tn we have∫

Tn∂ j f (x)g(x)dx =−

∫

Tn∂ jg(x) f (x)dx .

Elements of Zn are denoted by m = (m1, . . . ,mn). For m ∈ Zn, we define the total

size of m to be the number |m|= (m21 + · · ·+m2

n)1/2. Recall that for x = (x1, . . . ,xn)

and y = (y1, . . . ,yn) in Rn,

x · y = x1y1 + · · ·+ xnyn

denotes the usual dot product. Finally, for x ∈ Tn, |x| denotes the usual Euclidean

norm of x. If we identify Tn with [−1/2,1/2]n, then |x| can be interpreted as the

distance of the element x from the origin, and then we have 0 ≤ |x| ≤ √n/2 for all

x ∈ Tn.

Multi-indices are elements of (Z+ ∪0)n. For a multi-index α = (α1, . . . ,αn),we denote the partial derivative ∂α1

1 · · ·∂αnn f by ∂α f . The spaces C k(Tn) of con-

tinuously differentiable functions of order k, where k ∈ Z+, are defined as the sets

of functions ϕ for which ∂αϕ exist and are continuous for all |α| ≤ k. When k = 0

we set C 0(Tn) = C (Tn) to be the space of continuous functions on Tn. The space

C ∞(Tn) of infinitely differentiable functions on Tn is the union of all the C k(Tn).All of these spaces are contained in Lp(Tn), which are nested, with L1(Tn) being

the largest.

3.1.2 Fourier Coefficients

Definition 3.1.1. For a complex-valued function f in L1(Tn) and m in Zn, we define

f (m) =∫

Tnf (x)e−2πim·xdx . (3.1.4)

We call f (m) the mth Fourier coefficient of f . We note that f (ξ ) is not defined for

ξ ∈ Rn \Zn, since the function x → e−2πiξ ·x is not 1-periodic in any coordinate and

therefore not well defined on Tn. For a finite Borel measure µ on Tn and m ∈ Zn the

expression

µ(m) =∫

Tne−2πim·xdµ (3.1.5)

is called the mth Fourier coefficient of µ .


The Fourier series of f at x ∈ Tn is the series

∑m∈Zn

f (m)e2πim·x. (3.1.6)

It is not clear at present in which sense and for which x ∈ Tn (3.1.6) converges. The

study of convergence of Fourier series is the main topic of study in this chapter.

We quickly recall the notation we introduced in Chapter 2. We denote by f the

complex conjugate of the function f , by f the function f (x) = f (−x), and by τy( f )the function τy( f )(x) = f (x−y) for all y ∈ Tn. We mention some elementary prop-

erties of Fourier coefficients.

Proposition 3.1.2. Let f , g be in L1(Tn). Then for all m,k ∈ Zn, λ ∈C, y ∈ Tn, and

all multi-indices α we have

(1) f +g(m) = f (m)+ g(m) ,

(2) λ f (m) = λ f (m) ,

(3) f (m) = f (−m) ,

(4)f (m) = f (−m) ,

(5) τy( f )(m) = f (m)e−2πim·y ,

(6) (e2πik(·) f )(m) = f (m− k) ,

(7) f (0) =∫

Tnf (x)dx ,

(8) supm∈Zn

| f (m)| ≤∥∥ f

∥∥L1(Tn)

,

(9) f ∗g(m) = f (m)g(m) ,

(10) ∂α f (m) = (2πim)α f (m), whenever f ∈ C α .

Proof. The proof of properties (1)–(10) is rather easy and is left to the reader. We

only sketch the proof of (9). We have

f ∗g(m) =∫

Tn

∫

Tnf (x− y)g(y)e−2πim·(x−y)e−2πim·y dydx = f (m)g(m) ,

where the interchange of integrals is justified by the absolute convergence of the

integrals and Fubini’s theorem.

Remark 3.1.3. The Fourier coefficients have the following property. For a function

f1 on Tn1 and a function f2 on Tn2 , the tensor function

( f1⊗ f2)(x1,x2) = f1(x1) f2(x2)


is a periodic function on Tn1+n2 whose Fourier coefficients are

f1⊗ f2(m1,m2) = f1(m1) f2(m2) , (3.1.7)

for all m1 ∈ Zn1 and m2 ∈ Zn2 .

Definition 3.1.4. A trigonometric polynomial on Tn is a function of the form

P(x) = ∑m∈Zn

ame2πim·x, (3.1.8)

where amm∈Zn is a finitely supported sequence in Zn. The degree of P is the largest

number |q1|+ · · ·+ |qn| such that aq is nonzero, where q = (q1, . . . ,qn). Observe that

in view of the orthonormality of the exponentials we have for all m ∈ Zn

P(m) = am .

Example 3.1.5. If the sequence amm has only one nonzero term, then the trigono-

metric polynomial of Definition 3.1.4 reduces to a trigonometric monomial, which

has the form

P(x) = ae2πi(q1x1+···+qnxn)

for some q = (q1, . . . ,qn) ∈ Zn and a ∈ C.

Let

P(x) = ∑|m|≤N

ame2πim·x = ∑|m|≤N

P(m)e2πim·x

be a trigonometric polynomial on Tn and let µ be a finite Borel measure on Tn. Then

we have

(P∗µ)(x) =∫

Tn∑|m|≤N

P(m)e2πim·(x−y) dµ(y) = ∑|m|≤N

P(m)µ(m)e2πim·x. (3.1.9)

In particular, if f is an integrable function on Tn we have

(P∗ f )(x) =∫

Tnf (y) ∑

|m|≤N

P(m)e2πim·(x−y) dy = ∑|m|≤N

P(m) f (m)e2πim·x. (3.1.10)

This implies that the partial sums

∑|m|≤N

f (m)e2πim·x

of the Fourier series of f in (3.1.6) can be obtained by convolving f with the function

DN(x) = ∑|m|≤N

e2πim·x . (3.1.11)

This function is called the Dirichlet kernel.

3.1.3 The Dirichlet and Fejer Kernels

Definition 3.1.6. Let 0≤ R < ∞. The square Dirichlet kernel on Tn is the function

DnR(x) = ∑

m∈Zn

|m j |≤R

e2πim·x . (3.1.12)

The circular (or spherical) Dirichlet kernel on Tn is the function

Dn

R(x) = ∑m∈Zn

|m|≤R

e2πim·x . (3.1.13)

In dimension n = 1 these functions coincide and are denoted by

DR(x) = D1R(x) =

D1

R(x) .

This function is called the Dirichlet kernel and coincides with DN(x) in (3.1.11)

when N ≤ R < N +1 and N ∈ Z+∪0; see Figure 3.2.

-0.4 -0.2 0.2 0.4

-2

2

4

6

8

10

Fig. 3.2 The graph of the Dirichlet kernel D5 plotted on the interval [−1/2,1/2].

Both the square and circular (or spherical) Dirichlet kernels are trigonomet-

ric polynomials. The square Dirichlet kernel on Tn is equal to a product of one-

dimensional Dirichlet kernels, that is,

DnR(x1, . . . ,xn) = DR(x1) · · ·DR(xn) . (3.1.14)

We have the following two equivalent ways to write the Dirichlet kernel DN :

DN(x) = ∑|m|≤N

e2πim·x =sin((2N +1)πx)

sin(πx), (3.1.15)


for x ∈ [0,1]. To verify the validity of (3.1.15), we write

∑|m|≤N

e2πim·x = e−2πiNx e2πi(2N+1)x−1

e2πix−1=

e2πi(N+1)x− e−2πiNx

eπix(eπix− e−πix)=

sin((2N +1)πx)

sin(πx).

It follows that for R ∈ R+∪0 we have

DR(x) =sin(πx(2[R]+1))

sin(πx). (3.1.16)

It is reasonable to ask whether the family DRR>0 forms an approximate identity

as R → ∞. Using (3.1.15) we see that each DR is integrable over [−1/2,1/2] and

has integral equal to 1. But it follows from Exercise 3.1.5 that ‖DR‖L1 ≈ logR as

R → ∞, and therefore property (i) in Definition 1.2.15 fails for DR. We conclude

that the family DRR>0 is not an approximate identity on T1, which significantly

complicates the study of Fourier series. Consequently, the family DnRR>0 is not an

approximate identity on Tn, since ‖DnR‖L1(T1) ≈ (logR)n. The same is true for the

family of circular (or spherical) Dirichlet kernels DnRR>0. Although this is harder

to prove, it will be a consequence of the results in Section 4.2.

A typical situation encountered in analysis is that the means of a sequence behave

better than the original sequence. This fact led Cesaro and independently Fejer to

consider the arithmetic means of the Dirichlet kernel in dimension 1, that is, the

expressions

FN(x) =1

N +1

[D0(x)+D1(x)+D2(x)+ · · ·+DN(x)

]. (3.1.17)

The expression in (3.1.17) is in fact equal to the Fejer kernel given in Example 1.2.18.

We have the following identity concerning the kernel FN .

Proposition 3.1.7. For every nonnegative integer N the identity holds

FN(x) =N

∑j=−N

(1− | j|

N +1

)e2πi jx =

1

N +1

(sin(π(N +1)x)

sin(πx)

)2

(3.1.18)

for all x ∈ T1. Thus FN(m) = 1− |m|N+1

if |m| ≤ N and zero otherwise.

Proof. The fact that the expression in (3.1.17) is equal to the middle term in (3.1.18)

is a consequence of the trivial calculation:

1

N +1

N

∑k=0

Dk(x) =1

N +1

N

∑k=0∑| j|≤k

e2πi jx = ∑| j|≤N

#k ∈ Z : | j| ≤ k ≤ NN +1

e2πi jx .

To verify the second equality in (3.1.18) we use the simple geometric series identity

1+ r+ r2 + · · ·+ rN = 1−rN+1

1−rto write for x = 0


N

∑j=1

e2πi jx =e2πi(N+1)x−1

e2πix−1−1 =

eiπ(N+1)x

eiπx

eπi(N+1)x− e−πi(N+1)x

eπix− e−πix−1


N

∑j=1

| j|e2πi jx =1

2πi

d

dx

(eiπNx sin(π(N +1)x)

sin(πx)

). (3.1.19)

Likewise we prove that

−1

∑j=−N

| j|e2πi jx =− 1

2πi

d

dx

(e−iπNx sin(π(N +1)x)

sin(πx)

). (3.1.20)

Adding (3.1.19) and (3.1.20) we deduce

∑| j|≤N

| j|e2πi jx =1

π

d

dx

(sin(πNx)sin(π(N +1)x)

sin(πx)

). (3.1.21)

Multiplying (3.1.21) by − 1N+1

and adding DN(x) we obtain

N

∑j=−N

(1− | j|

N +1

)e2πi jx =

sin(π(2N +1)x)

sin(πx)− 1

π

d

dx

(sin(πNx)

sin(π(N +1)x)

(N +1)sin(πx)

).

Writing the preceding expression on the right as

(N+1)sin(π(N+1)x)cos(πNx)sin(πx)+(N+1)cos(π(N+1)x)sin(πNx)sin(πx)

(N+1)sin2(πx)

− 1

π

ddx

sin(πNx)sin(π(N+1)x)

sin(πx)−

sin(πNx)sin(π(N+1)x)

π cos(πx)

(N+1)sin2(πx),

computing the derivative of the expression in the curly brackets, and simplifying,

we finally obtain that

N

∑j=−N

(1− | j|

N +1

)e2πi jx =

1

N +1

(sin(π(N +1)x)

sin(πx)

)2

. (3.1.22)

This proves the second identity in (3.1.18).

Definition 3.1.8. Let N be a nonnegative integer. The function FN on T1 given by

(3.1.22) is called the Fejer kernel.

The Fejer kernel FnN on Tn is defined as the product of the 1-dimensional Fejer

kernels, or as the average of the product of the Dirichlet kernels in each variable,

precisely, F1N(x) = FN(x) and


FnN(x1, . . . ,xn) =

n

∏j=1

FN(x j)

=n

∏j=1

(1

N +1

N

∑k=0

Dk(x j)

)

=1

(N +1)n

N

∑k1=0

. . .N

∑kn=0

Dk1(x1) · · ·Dkn

(xn) .

Note that FnN is a trigonometric polynomial of degree nN.

Remark 3.1.9. Using the identities for FN in (3.1.18), we may write for all N ≥ 0

FnN(x1, . . . ,xn) = ∑

m∈Zn

|m j |≤N

(1− |m1|

N +1

)· · ·

(1− |mn|

N +1

)e2πim·x (3.1.23)

=1

(N +1)n

n

∏j=1

(sin(π(N +1)x j)

sin(πx j)

)2

, (3.1.24)

thus FnN ≥ 0. Observe that Fn

0 (x) = 1 for all x ∈ Tn and that FnN(0) = (N +1)n.

Proposition 3.1.10. The family of Fejer kernels FnN∞N=0 is an approximate identity

on Tn.

Proof. Since FnN ≥ 0 we have that ‖Fn

N‖L1 =∫

Tn FnN dx. Also

∫Tn Fn

N dx = 1, in view

of identity (3.1.23). Thus properties (i) and (ii) of approximate identities (according

to Definition 1.2.15) hold. To prove property (iii) of the definition we make use of

identity (3.1.24). Using the fact that 1≤ |t||sin t| ≤

π2

when |t| ≤ π2

, we obtain

FN(x)≤1

N +1min

((N +1)|πx||sin(πx)| ,

1

|sin(πx)|

)2

≤ 1

N +1

π2

4min

(N +1,

1

|πx|

)2

when |x| ≤ 12. This implies that for δ > 0 we have

∫

δ≤|x|≤ 12

FN(x)dx≤ 1

N +1

π2

4

∫

δ≤|x|≤ 12

dx

|πδ |2 ≤1

4δ 2

1

N +1→ 0

as N →∞. In higher dimensions, given x = (x1, . . . ,xn) ∈ [−1/2,1/2]n with |x| ≥ δ ,

there is a j ∈ 1, . . . ,n such that |x j| ≥ δ/√

n and thus

∫x∈Tn

|x|≥δFn

N(x)dx≤n

∑j=1

∫

|x j |≥ δ√n

FN(x j)dx j∏k = j

∫

T1FN(xk)dxk ≤

n

4(δ/√

n)2

1

N +1→ 0 .

This proves the claim.


Exercises

3.1.1. Identify T1 with [−1/2,1/2] and let h(t) be an integrable function on T1.

(a) If h(t)≥ 0 is even, show that h(m) is real and |h(m)| ≤ h(0) for all m ∈ Z.

(b) If h(t) is odd and h(t) ≥ 0 on [0,1/2), then i h(m) is real and |h(m)| ≤ imh(1)for all m ∈ Z.

3.1.2. Suppose that h is a periodic integrable function on [−1/2,1/2) with integral

zero. Define another periodic function H on T1 by setting

H(x) =∫ x

−1/2h(t)dt .

Compute H(m) in terms of h(m) for m ∈ Z.

3.1.3. Suppose that gεε>0 is an approximate identity on Rn as ε → 0 and let

Gε(x) = ∑ℓ∈Zn

gε(x+ ℓ) .

Show that the family Gεε>0 is an approximate identity on Tn.

3.1.4. On T1 define the de la Vallee Poussin kernel

VN(x) = 2F2N+1(x)−FN(x) .

(a) Show that the sequence VN is an approximate identity.

(b) Prove that VN(m) = 1 when |m| ≤ N +1, and VN(m) = 0 when |m| ≥ 2N +2.

3.1.5. (a) Show that for all |t| ≤ π2

we have

∣∣∣ 1

sin(t)− 1

t

∣∣∣≤ 1− 2

π.

(b) Let DN be the Dirichlet kernel on T1. Prove that for N ∈ Z+ we have

4

π2

N

∑k=1

1

k≤

∥∥DN

∥∥L1 ≤ 3− 2

π+

4

π2

N

∑k=1

1

k.

Conclude that the numbers ‖DN‖L1 grow logarithmically as N → ∞ and therefore

the family DN∞N=1 is not an approximate identity on T1. The numbers ‖DN‖L1 ,

N = 1,2, . . . are called the Lebesgue constants.[Hint: Part (a): Show that the derivative of 1

sin(t) −1t

is nonnegative on (0, π2], or

equivalently prove that tan(t)sin(t) ≥ t2 on (0, π2]; this is a consequence of the in-

equality√

sin(t) tan(t)≥ 2( 1sin(t) +

1tan(t) )

−1 = 2tan( t2)≥ t. Part (b): Replace DN(t)

bysin((2N+1)πt)

πtand estimate the difference using part (a).

]

3.2 Reproduction of Functions from Their Fourier Coefficients 183

3.1.6. Let DN be the Dirichlet kernel on T1. Prove that for all 1 0 such that

cp (2N +1)1/p′ ≤∥∥DN

∥∥Lp ≤Cp (2N +1)1/p′ .

[Hint: Consider the two closest zeros of DN near the origin and split the integral

into the intervals thus obtained.]

3.1.7. The Poisson kernel on Tn is the function

Pr1,...,rn(x) = ∑m∈Zn

r|m1|1 · · ·r|mn|

n e2πim·x

and is defined for 0 < r1, . . . ,rn < 1. Prove that Pr1,...,rn can be written as

Pr1,...,rn(x1, . . . ,xn) =n

∏j=1

Re

(1+ r je

2πix j

1− r je2πix j

)=

n

∏j=1

1− r2j

1−2r j cos(2πx j)+ r2j

,

and conclude that Pr,...,r(x) is an approximate identity as r ↑ 1.

3.2 Reproduction of Functions from Their Fourier Coefficients

We can obtain very interesting results using the Fejer kernel.

Proposition 3.2.1. The set of trigonometric polynomials is dense in Lp(Tn) for 1≤p < ∞.

Proof. Given f in Lp(Tn) for 1 ≤ p < ∞, consider f ∗FnN . Clearly f ∗Fn

N is also a

trigonometric polynomial. In view of Theorem 1.2.19 (1), f ∗FnN converges to f in

Lp as N → ∞.

Corollary 3.2.2. (Weierstrass approximation theorem for trigonometric polyno-

mials) Every continuous function on the torus is a uniform limit of trigonometric

polynomials.

Proof. Since f is continuous on Tn, which is a compact set, Theorem 1.2.19 (2)

gives that f ∗FnN converges uniformly to f as N → ∞. But f ∗Fn

N is a trigonomet-

ric polynomial, and so we conclude that every continuous function on Tn can be

uniformly approximated by trigonometric polynomials.

3.2.1 Partial sums and Fourier inversion

We now define the partial sums of Fourier series.


Definition 3.2.3. Let R≥ 0 and N ∈ Z+∪0. The expressions

( f ∗DnR)(x) = ∑

m∈Zn

|m j |≤R

f (m)e2πim·x

are called the square partial sums of the Fourier series of f . Then the expressions

( f ∗Dn

R)(x) = ∑m∈Zn

|m|≤R

f (m)e2πim·x

are called the circular (or spherical) partial sums of the Fourier series of f . The

expressions

( f ∗FnN)(x) = ∑

m∈Zn

|m j |≤N

(1− |m1|

N +1

)· · ·

(1− |mn|

N +1

)f (m)e2πim·x

are called the square Cesaro means (or square Fejer means) of f .

Finally, for R≥ 0 the expressions

( f ∗Fn

R )(x) = ∑m∈Zn

|m|≤R

(1− |m|

R

)f (m)e2πim·x

are called the circular Cesaro means (or circular Fejer means) of f .

Observe that f ∗ Fn

R is equal to the average of the expressions f ∗ Dn

r in the

following sense:

( f ∗Fn

R )(x) =1

R

∫ R

0( f ∗

Dn

r )(x)dr.

This is analogous to the fact that the Fejer kernel FN is the average of the Dirichlet

kernels D0, D1, . . . ,DN .

A fundamental problem is in what sense the partial sums of the Fourier series

converge back to the function as R→ ∞ or N → ∞. This problem is of central im-

portance in harmonic analysis and is in part investigated in this chapter.

We now ask the question whether the Fourier coefficients uniquely determine the

function. The answer is affirmative and simple.

Proposition 3.2.4. If f ,g ∈ L1(Tn) satisfy f (m) = g(m) for all m in Zn, then

f = g a.e.

Proof. By linearity of the problem, it suffices to assume that g = 0. If f (m) = 0

for all m ∈ Zn, Definition 3.2.3 implies that FnN ∗ f = 0 for all N ∈ Z+. In view of

Proposition 3.1.10, the sequence FnNN∈Z+ is an approximate identity as N → ∞.

Therefore, ∥∥ f −FnN ∗ f

∥∥L1 → 0

as N → ∞; hence ‖ f‖L1 = 0, from which we conclude that f = 0 a.e.

A useful consequence of the result just proved is the following.

Proposition 3.2.5. (Fourier inversion) Suppose that f ∈ L1(Tn) and that

∑m∈Zn

| f (m)|< ∞ .

Then

f (x) = ∑m∈Zn

f (m)e2πim·x a.e., (3.2.1)

and therefore f is almost everywhere equal to a continuous function.

Proof. It is straightforward to check that both functions in (3.2.1) are well defined

and have the same Fourier coefficients. Therefore, they must be almost everywhere

equal by Proposition 3.2.4. Moreover, the function on the right in (3.2.1) is every-

where continuous.

3.2.2 Fourier series of square summable functions

Let H be a separable Hilbert space with complex inner product 〈· | ·〉. Recall that

a subset E of H is called orthonormal if 〈 f |g〉 = 0 for all f , g in E with f = g,

while 〈 f | f 〉 = 1 for all f in E. A complete orthonormal system is a subset of H

having the additional property that the only vector orthogonal to all of its elements

is the zero vector. We summarize basic properties about orthonormal systems in the

proposition below (see [307]).

Proposition 3.2.6. Let H be a separable Hilbert space and let ϕkk∈Z be an or-

thonormal system in H. Then the following are equivalent:

(1) ϕkk∈Z is a complete orthonormal system.

(2) For every f ∈ H we have

∥∥ f∥∥2

H= ∑

k∈Z

|〈 f |ϕk〉|2 .

(3) For every f ∈ H we have

f = limN→∞

∑|k|≤N

〈 f |ϕk〉ϕk ,

where the series converges in H.

Now consider the Hilbert space space L2(Tn) with inner product

⟨f |g

⟩=

∫

Tnf (t)g(t)dt .


Let ϕm be the sequence of functions ξ → e2πim·ξ indexed by m ∈ Zn. The orthonor-

mality of the sequence ϕm is a consequence of the following simple but crucial

identity:∫

[0,1]ne2πim·x e2πik·x dx =

1 when m = k,

0 when m = k.

The completeness of the sequence ϕm is also evident. Since⟨

f |ϕm

⟩= f (m) for

all f ∈ L2(Tn), it follows from Proposition 3.2.4 that if⟨

f |ϕm

⟩= 0 for all m ∈ Zn,

then f = 0 a.e.

The next result is a consequence of Proposition 3.2.6.

Proposition 3.2.7. The following are valid for f ,g ∈ L2(Tn):(1) (Plancherel’s identity)

∥∥ f∥∥2

L2 = ∑m∈Zn

| f (m)|2 .

(2) The function f (t) is a.e. equal to the L2(Tn) limit of the sequence

limM→∞

∑|m|≤M

f (m)e2πim·t .

(3) (Parseval’s relation)

∫

Tnf (t)g(t)dt = ∑

m∈Zn

f (m)g(m) .

(4) The map f → f (m)m∈Zn is an isometry from L2(Tn) onto ℓ2.

(5) For all k ∈ Zn we have

f g(k) = ∑m∈Zn

f (m)g(k−m) = ∑m∈Zn

f (k−m)g(m) .

Proof. (1) and (2) follow from the corresponding statements in Proposition 3.2.6.

Notice that both sides of (3) converge by the Cauchy-Schwarz inequality. Parseval’s

relation (3) follows from polarization. By this we mean the following procedure.

First replace f by f +g in (1) and expand the squares to obtain

‖ f‖2L2 +‖g‖2

L2 +2Re〈 f |g〉=‖ f +g‖2L2

= ∑m∈Zn

| f (m)+ g(m)|2

= ∑m∈Zn

| f (m)|2 + ∑m∈Zn

|g(m)|2 +2Re ∑m∈Zn

f (m)g(m)

and from this it follows that the real parts of the expressions in (3) are equal. Next

replace f by f + ig in (1) and expand the squares. Using Re(−iw) = Imw we obtain

‖ f‖2L2 +‖g‖2

L2 +2Im〈 f |g〉=‖ f + ig‖2L2

= ∑m∈Zn

| f (m)+ i g(m)|2

= ∑m∈Zn

| f (m)|2 + ∑m∈Zn

|g(m)|2 +2Im ∑m∈Zn

f (m)g(m),

and thus the imaginary parts of the expressions in (3) are equal. Thus (3) holds.

Next we prove (4). We already know that the map f → f (m)m∈Zn is an injective

isometry. It remains to show that it is onto. Given a square summable sequence

amm∈Zn of complex numbers, define

fN(t) = ∑|m|≤N

ame2πim·t .

Observe that fN is a Cauchy sequence in L2(Tn) and it therefore converges to some

f ∈ L2(Tn). Then we have f (m) = am for all m ∈ Zn. Finally, (5) is a consequence

of (3) and Proposition 3.1.2 (6) and (3).

3.2.3 The Poisson Summation Formula

We end this section with an important result that connects Fourier analysis on the

torus with Fourier analysis on Rn. Suppose that f is an integrable function on Rn

and let f be its Fourier transform. Restrict f on Zn and form the “Fourier series”

(assuming that it converges)

∑m∈Zn

f (m)e2πim·x.

What does this series represent? Since the preceding function is 1-periodic in every

variable, it follows that it cannot be equal to f , unless f is itself periodic. However,

it should not come as a surprise that it is in fact equal to the periodization of f

on Rn. In other words, the Fourier expansion of a function on Rn reproduces the

periodization of the function.

Theorem 3.2.8. (Poisson summation formula) Let f be a continuous function on

Rn which satisfies for some C,δ > 0 and for all x ∈ Rn

| f (x)| ≤C(1+ |x|)−n−δ ,

and whose Fourier transform f restricted on Zn satisfies

∑m∈Zn

| f (m)|< ∞ . (3.2.2)

Then for all x ∈ Rn we have

∑m∈Zn

f (m)e2πim·x = ∑k∈Zn

f (x+ k), (3.2.3)

and in particular

∑m∈Zn

f (m) = ∑k∈Zn

f (k).

Proof. Define a 1-periodic function on Tn by setting

F(x) = ∑k∈Zn

f (x+ k) .

It is straightforward to verify that ‖F‖L1([0,1]n) = ‖ f‖L1(Rn), thus F lies in L1(Tn). We

prove that the sequence of the Fourier coefficients of F coincides with the restriction

of the Fourier transform of f on Zn. This follows from the calculation

F(m) =∫

Tn∑

k∈Zn

f (x+ k)e−2πim·x dx

= ∑k∈Zn

∫

Tnf (x+ k)e−2πim·x dx

= ∑k∈Zn

∫

[− 12 ,

12 ]

n−kf (x)e−2πim·x dx

=∫

Rnf (x)e−2πim·x dx

= f (m) ,

in which the interchange of the sum and the integral is justified by the Weierstrass

M-test of uniform convergence of series, since

∑k∈Zn

1

(1+ |k+ x|)n+δ≤ ∑

k∈Zn

(1+√

n)n+δ

(1+√

n+ |k+ x|)n+δ≤ ∑

k∈Zn

Cn,δ

(1+ |k|)n+δ< ∞ ,

where we used |k+ x| ≥ |k|− |x| ≥ |k|−√n. This calculation also shows that F is

the sum of a uniformly convergent series of continuous functions on [0,1]n, thus

it is itself continuous. It follows that (3.2.2) holds with |F(m)| in place of | f (m)|.Hence, Proposition 3.2.5 applies, and given the fact that F is continuous, it yields

conclusion (3.2.3) for all x ∈ Tn and, by periodicity, for all x ∈ Rn.

Example 3.2.9. We have seen earlier (Exercise 2.2.11) that the following identity

gives the Fourier transform of the Poisson kernel in Rn:

(e−2π|x|)(ξ ) = Γ ( n+12)

πn+1

2

1

(1+ |ξ |2) n+12

.

The Poisson summation formula yields the identity

Γ ( n+12)

πn+1

2∑

k∈Zn

ε−n

(1+ |k+x|2ε2 )

n+12

= ∑m∈Zn

e−2πε |m|e−2πim·x (3.2.4)

which implies

Γ ( n+12)

πn+1

2∑

k∈Zn

1

(1+ |k|2) n+12

= ∑m∈Zn

e−2π|m| . (3.2.5)


∑k∈Zn\0

1

(ε2 + |k|2) n+12

=1

ε

(π

n+12

Γ ( n+12)∑

m∈Zn

e−2πε |m|− 1

εn

),

from which we obtain the identity

∑k∈Zn\0

1

|k|n+1= lim

ε→0

1

ε

(π

n+12

Γ ( n+12)∑

m∈Zn

e−2πε |m|− 1

εn

). (3.2.6)

The limit in (3.2.6) can be easily calculated in dimension 1 using that

limδ→0

π2

δ

(1+ e−2δ

1− e−2δ− 1

δ

)=π2

3,

and this yields

∑k =0

1

k2=π2

3.

Also, in dimension 1, from (3.2.5) we obtain the related identity

∑k∈Z

1

1+ k2= π ∑

m∈Z

e−2π|m| = π1+ e−2π

1− e−2π.

Example 3.2.10. Let 0 < Re α < n. We introduce a smooth function Φ(ξ ) which is

equal to 1 on the ball |ξ | ≤ 1 and vanishes outside the ball |ξ | ≤ 2. We investigate

the behavior as x→ 0 of the expression

limR→∞

∑m∈Zn\0

e2πim·x

|m|α Φ(m

R

)

when x ∈ [− 12, 1

2)n \0. As in Example 2.4.9, let η be a smooth radial function on

Rn that is equal to 1 outside the ball B(0,1/2) and vanishes on the ball B(0,1/4)and define

g = (η(ξ )|ξ |−α) .


Let Φδ (x) = δ−nΦ(x/δ ). The Poisson summation formula (Theorem 3.2.8) gives

∑k∈Zn\0

e2πik·x

|k|α Φ

(k

R

)= ∑

k∈Zn

η(k)e2πik·x

|k|α Φ

(k

R

)

= (g∗Φ1/R)(x)+ ∑m∈Zn\0

(g∗Φ1/R)(x+m) .

It was shown in Example 2.4.9 that g(y) decays faster than the reciprocal of any

polynomial at infinity and is equal to πα−n2Γ ( n−α

2)Γ (α

2)−1|y|α−n + h(y), where h

is a smooth function on Rn. Since x = 0, it follows that g is smooth in a small rela-

tively compact neighborhood of x and, since Φ1/RR>0 is an approximate identity,

Theorem 1.2.19 (2) yields that (g∗Φ1/R)(x)→ g(x) as R→∞. Assume for a moment

that

limR→∞

∑m∈Zn\0

(g∗Φ1/R)(x+m) = ∑m∈Zn\0

limR→∞

(g∗Φ1/R)(x+m) . (3.2.7)

Since x+m does not vanish for any m ∈ Zn \ 0, the function g is smooth in a

relatively compact neighborhood of x+m, and thus (g ∗Φ1/R)(x+m)→ g(x+m)as R→ ∞. Consequently, the sum on the right in (3.2.7) is equal to

∑m∈Zn\0

g(x+m) .

We conclude that

limR→∞

∑m∈Zn\0

e2πim·x

|m|α Φ(m

R

)=πα−

n2Γ ( n−α

2)

Γ (α2)

|x|α−n +h1(x) ,

where h1 is a C ∞ function on [− 12, 1

2)n given by

h1(t) = h(t)+ ∑m∈Zn\0

g(t +m) .

We now explain the passage of the limit inside the sum in (3.2.7). This is a con-

sequence of the Lebesgue dominated convergence theorem, provided we know that

|(g∗Φ1/R)(x+m)| ≤ C

|m|n+1, |m|> 5

√n (3.2.8)

for some constant C independent of R ≥ 1 and of m. Indeed, the expression on the

left of this inequality is bounded by I + II, where

I = Cn,α

∫

|x+m−y|≤2√

n|x+m− y|α−n Rn

(1+R|y|)2n+2dy

II = Cn,α

∫

|x+m−y|≥2√

n

1

(1+ |x+m− y|)2n+2

Rn

(1+R|y|)2n+2dy .

In I we have

1+R|y| ≥ R|x+m|−R|x+m− y| ≥ R|m|− 12R√

n−2R√

n≥ 12R|m| ,

hence

I ≤C′n,αR−n−2|m|−2n−2 ≤C′n,α |m|−2n−2 .

In II we use

(1+ |x+m− y|)n+1(1+R|y|)n+1 ≥ (1+ |m|)n+1

while the term left produces a convergent integral, which is uniformly bounded in

R≥ 1. This proves (3.2.8).

Exercises

3.2.1. On T1 let P be a trigonometric polynomial of degree N > 0. Show that P has

at most 2N zeros. Construct a trigonometric polynomial with exactly 2N zeros.

3.2.2. (Hausdorff–Young inequality) Prove that when f ∈ Lp(Tn), 1 ≤ p ≤ 2, the

sequence of Fourier coefficients of f is in ℓp′(Zn) and

(∑

m∈Zn

| f (m)|p′)1/p′

≤∥∥ f

∥∥Lp(Tn)

.

Also observe that 1 is the best constant in the preceding inequality.

3.2.3. Use without proof that there exists a constant C > 0 such that for all t ∈R we

have ∣∣∣∣N

∑k=2

eik logkeikt

∣∣∣∣≤C√

N, N = 2,3,4, . . . ,

to construct an example of a continuous function g on T1 with

∑m∈Z

|g(m)|q = ∞

for all q < 2. Thus the Hausdorff–Young inequality of Exercise 3.2.2 fails for p > 2.[Hint: Consider g(x) = ∑∞k=2

eiklogk

k1/2(logk)2e2πikx. For a proof of the previous estimate,

see Zygmund [388, Theorem (4.7) p. 199].]

3.2.4. (S. Bernstein) Let P(x) be a trigonometric polynomial of degree N on T1.

Prove that ‖P′‖L∞ ≤ 4πN‖P‖L∞ .[Hint: Prove first that P′(x)/2πiN is equal to

((e−2πiN(·)P)∗FN−1

)(x)e2πiNx−

((e2πiN(·)P)∗FN−1

)(x)e−2πiNx

and then take L∞ norms.]

3.2.5. (Fejer and F. Riesz) Let P(ξ ) = ∑Nk=−N ake2πikξ be a trigonometric polyno-

mial on T1 of degree N such that P(ξ )> 0 for all ξ . Prove that there exists a trigono-

metric polynomial Q(ξ ) of the form ∑Nk=0 bke2πikξ such that P(ξ ) = |Q(ξ )|2.[

Hint: Since P ≥ 0 the complex-variable polynomial R(z) = ∑Nk=−N akzk+N must

satisfy R(z) = z2NR(1/z), and thus it must have N zeros inside the unit circle and

the other N outside. Therefore we may write R(z) = aN ∏sk=1(z− zk)

rk(z− 1/zk)rk

for some 0 < |zk|< 1 and rk ≥ 1 with ∑sk=1 rk = N. Then take z = e2πiξ .

]

3.2.6. Let g be a function on Rn that satisfies |g(x)|+ |g(x)| ≤ C(1+ |x|)−n−δ for

some C,δ > 0 and all x ∈ Rn. Prove that

λ n ∑m∈Zn

g(λm+α)e2πix·(m+ αλ) = ∑

k∈Zn

g(x+ k

λ

)e−2πi k·α

λ

for any x,α ∈ Rn and λ > 0.

3.2.7. Verify the following identity when 0 < r < 1 and x ∈ Rn

Γ ( n+12)

πn+1

2∑

k∈Zn

12π log 1

r(( 1

2π log 1r)2 + |x− k|2

) n+12

= ∑m∈Zn

r|m|e2πim·x .

In the special case n = 1 and x ∈ R we have

1

π ∑k∈Z

12π log 1

r

( 12π log 1

r)2 + |x− k|2

=1− r2

1−2r cos(2πx)+ r2.

[Hint: Use identity (3.2.4) and Exercise 3.1.7 when n = 1.

]

3.2.8. Let γ ∈ R and λ > 0. Show that

∑k∈Z

cos(2πkγ)

λ 2 + k2=π

λ

cosh(2πλ (γ− [γ ]− 12))

sinh(πλ ).

[Hint: Use Exercise 3.2.6 when n = 1 with x = 0, α =−γλ , g(x) = 1

π1

1+x2 and sum

in m.]

3.3 Decay of Fourier Coefficients

In this section we investigate the interplay between the smoothness of a function and

the decay of its Fourier coefficients.

3.3 Decay of Fourier Coefficients 193

3.3.1 Decay of Fourier Coefficients of Arbitrary Integrable

Functions

We begin with the classical result asserting that the Fourier coefficients of any inte-

grable function tend to zero at infinity. One should compare the following proposi-

tion with Proposition 2.2.17.

Proposition 3.3.1. (Riemann–Lebesgue lemma) Given a function f in L1(Tn), we

have that | f (m)| → 0 as |m| → ∞.

Proof. Given f ∈ L1(Tn) and ε > 0, let P be a trigonometric polynomial such that

‖ f −P‖L1 < ε . If |m|> degree(P), then P(m) = 0 and thus

| f (m)|= | f (m)− P(m)| ≤∥∥ f −P

∥∥L1 < ε .

This proves that | f (m)| → 0 as |m| → ∞.

Several questions are naturally raised. How fast can the Fourier coefficients of an

L1 function tend to zero? Does additional smoothness of the function imply faster

decay of the Fourier coefficients? Can such a decay be quantitatively expressed in

terms of the smoothness of the function?

We answer the first question. Fourier coefficients of an L1 function may tend to

zero arbitrarily slowly, that is, slower than any given rate of decay. To achieve this,

we need the following two lemmas.

Lemma 3.3.2. Given a sequence of positive real numbers am∞m=0 that tends to

zero as m→ ∞, there exists a sequence cm∞m=0 that satisfies

cm ≥ am, cm ↓ 0, and cm+2 + cm ≥ 2cm+1 (3.3.1)

for all m = 0,1, . . . . A sequence cm∞m=0 that satisfies (3.3.1) is called convex.

Proof. Let k0 = 0 and suppose that am ≤ M for all m ≥ 0. Find k1 > k0 such that

for m ≥ k1 we have am ≤ M/2. Now find k2 > k1 +k1−k0

2such that for m ≥ k2 we

have am ≤M/4. Next find k3 > k2 +k2−k1

2such that for m≥ k3 we have am ≤M/8.

Continue inductively in this way and construct a subsequence k0 < k1 < k2 < · · · of

the integers such that for m ≥ k j+1 we have am ≤ 2− j−1M and k j+1 > k j +k j−k j−1

2

for j ≥ 1. Join the points (k0,2M), (k1,M), (k2,M/2), (k3,M/4), . . . by straight

lines and note that by the choice of the sequence k j∞j=0 the resulting piecewise

linear function h is convex on [0,∞). Define cm = h(m) and observe that the se-

quence cm∞m=0 satisfies the required properties. Exercise 3.3.1 contains an alterna-

tive proof.

Lemma 3.3.3. Given a convex decreasing sequence cm∞m=0 of positive real num-

bers satisfying limm→∞ cm = 0 and a fixed integer s≥ 0, we have that

∞

∑r=0

(r+1)(cr+s + cr+s+2−2cr+s+1) = cs . (3.3.2)

Proof. We begin by observing the validity of the telescoping sum

N

∑r=0

(r+1)(cr+s + cr+s+2−2cr+s+1)

= cs− (N +1)(cs+N+1− cs+N+2)− cs+N+1 .

(3.3.3)

To show that the last expression tends to cs as N → ∞, we take M = [N2] and we use

convexity(cs+M+ j−cs+M+ j+1 ≥ cs+M+ j+1−cs+M+ j+2

)to obtain

cs+M+1− cs+N+2 = cs+M+1− cs+M+2

+ cs+M+2− cs+M+3

+ · · ·+ cs+N+1− cs+N+2

≥ (N−M+1)(cs+N+1− cs+N+2)

≥ N+12

(cs+N+1− cs+N+2)≥ 0 .

The preceding calculation implies that (N + 1)(cs+N+1− cs+N+2) tends to zero as

N → ∞ and thus the expression in (3.3.3) converges to cs as N → ∞.

The proof of Lemma 3.3.3 appears more natural after examining Exercise 3.3.3(a).

We now state the theorem we alluded to earlier.

Theorem 3.3.4. Let (dm)m∈Zn be a sequence of positive real numbers with dm → 0

as |m| → ∞. Then there exists a function f ∈ L1(Tn) such that f (m) ≥ dm for all

m ∈ Zn. In other words, given any rate of decay, there exists an integrable function

on the torus whose Fourier coefficients have slower rate of decay.

Proof. We are given a sequence of positive numbers amm∈Z that converges to zero

as |m| → ∞ and we would like to find an integrable function on T1 with f (m)≥ am

for all m ∈ Z. Apply Lemma 3.3.2 to the sequence am +a−mm≥0 to find a convex

sequence cmm≥0 that dominates am +a−mm≥0 and decreases to zero as m→ ∞.

Extend cm for m < 0 by setting cm = c|m|. Now define

f (x) =∞

∑j=0

( j+1)(c j + c j+2−2c j+1)Fj(x) , (3.3.4)

where Fj is the (one-dimensional) Fejer kernel. The convexity of the sequence cm

and the positivity of the Fejer kernel imply that f ≥ 0. Lemma 3.3.3 with s = 0 gives

that∞

∑j=0

( j+1)(c j + c j+2−2c j+1)∥∥Fj

∥∥L1 = c0 < ∞ , (3.3.5)

since ‖Fj‖L1 = 1 for all j. Therefore (3.3.4) defines an integrable function f on T1.

We now compute the Fourier coefficients of f . Since the series in (3.3.4) converges

in L1, for m ∈ Z we have

f (m) =∞

∑j=0

( j+1)(c j + c j+2−2c j+1)Fj(m)

=∞

∑j=|m|

( j+1)(c j + c j+2−2c j+1)

(1− |m|

j+1

)

=∞

∑r=0

(r+1)(cr+|m|+ cr+|m|+2−2cr+|m|+1)

= c|m| = cm ,

(3.3.6)

where we used Lemma 3.3.3 with s = |m|.Let us now extend this result on Tn. Let (dm)m∈Zn be a positive sequence with

dm → 0 as |m| →∞. By Exercise 3.3.2, there exists a positive sequence (a j) j∈Z with

am1· · ·amn ≥ d(m1,...,mn) and a j → 0 as | j| → ∞. Let

f(x1, . . . ,xn) = f (x1) · · · f (xn),

where f is the function previously constructed when n = 1 so that f (m)≥ am. It can

be seen easily using (3.1.7) that f(m)≥ dm.

3.3.2 Decay of Fourier Coefficients of Smooth Functions

We next study the decay of the Fourier coefficients of functions that possess a cer-

tain amount of smoothness. In this section we see that the decay of the Fourier

coefficients reflects the smoothness of the function in a rather precise quantitative

way. Conversely, one can infer some information about the smoothness of a function

from the decay of its Fourier coefficients.

Definition 3.3.5. Given 0 < γ < 1 and f a function on Tn, define the homogeneous

Lipschitz seminorm of order γ of f by

∥∥ f∥∥ .Λγ

= supx,h∈Tn

h =0

| f (x+h)− f (x)||h|γ

and define the homogeneous Lipschitz space of order γ as

.Λγ(T

n) = f : Tn → C with∥∥ f

∥∥ .Λγ

< ∞.

Functions in.Λγ(T

n) are called homogeneous Lipschitz functions of order γ .

There is an analogous definition for the inhomogeneous norm.

Definition 3.3.6. For 0 < γ < 1 and f a function on Tn, define the inhomogeneous

Lipschitz norm of order γ of f by

∥∥ f∥∥Λγ

=∥∥ f

∥∥L∞

+ supx,h∈Tn

h =0

| f (x+h)− f (x)||h|γ =

∥∥ f∥∥

L∞+

∥∥ f∥∥ .Λγ

.

Also define the inhomogeneous Lipschitz space of order γ as

Λγ(Tn) = f : Tn → C with

∥∥ f∥∥Λγ

< ∞.

Functions in Λγ(Tn) are called inhomogeneous Lipschitz functions of order γ .

Remark 3.3.7. Functions in both spaces Λγ(Tn) and

.Λγ(T

n) are obviously contin-

uous and therefore bounded. Moreover, the functional ‖ · ‖Λγ is a norm on Λγ(Tn).

The positive functional ‖ · ‖ .Λγ

satisfies the triangle inequality, but it does not satisfy

the property ‖ f‖ .Λγ

= 0 =⇒ f = 0 required to be a norm. It is therefore a semi-

norm on.Λγ(T

n). However, if we identify functions whose difference is a constant,

we form a space of the equivalence classes.Λγ(T

n)/constants on which ‖ · ‖ .Λγ

is

a norm.

Remark 3.3.8. We already observed that elements of.Λγ(T

n) are continuous and

thus bounded. Therefore,.Λγ(T

n)⊆ L∞(Tn) in the set-theoretic sense. However, the

norm inequality ‖ f‖L∞ ≤C‖ f‖ .Λγ

for all f ∈.Λγ fails for all constants C. For exam-

ple, take f = N + sin(2πx1) on Tn and let N → ∞ to see that this is the case.

The following theorem indicates how the smoothness of a function is reflected

by the decay of its Fourier coefficients.

Theorem 3.3.9. Let s ∈ Z with s≥ 0.

(a) Suppose that ∂α f exist and are integrable for all |α| ≤ s. Then

| f (m)| ≤(√

n

2π

)s max|α |=s

|∂α f (m)|

|m|s , m = 0, (3.3.7)

and thus

| f (m)|(1+ |m|s)→ 0

as |m| → ∞. In particular this holds when f lies in C s(Tn).(b) Suppose that ∂α f exist for all |α| ≤ s and whenever |α|= s, ∂α f are in

.Λγ(T

n)for some 0 < γ < 1. Then

| f (m)| ≤ (√

n)s+γ

(2π)s2γ+1

max|α |=s

∥∥∂α f∥∥ .Λγ

|m|s+γ , m = 0. (3.3.8)

Proof. Fix m ∈ Zn \ 0 and pick a j such that |m j| = sup1≤k≤n |mk|. Then clearly

m j = 0. Integrating by parts s times with respect to the variable x j, we obtain

f (m) =∫

Tnf (x)e−2πix·m dx = (−1)s

∫

Tn(∂ s

j f )(x)e−2πix·m

(−2πim j)sdx , (3.3.9)

where the boundary terms all vanish because of the periodicity of the integrand.

Taking absolute values and using |m| ≤ √n |m j|, we obtain assertion (3.3.7).

We now turn to the second part of the theorem. Let e j = (0, . . . ,1, . . . ,0) be the

element of the torus Tn whose jth coordinate is one and all the others are zero. A

simple change of variables together with the fact that eπi =−1 gives that

∫

Tn(∂ s

j f )(x)e−2πix·m dx =−∫

Tn(∂ s

j f )(x− e j

2m j)e−2πix·m dx ,

which implies that

∫

Tn(∂ s

j f )(x)e−2πix·m dx =1

2

∫

Tn

[(∂ s

j f )(x)− (∂ sj f )(x− e j

2m j)]e−2πix·m dx .

Now use the estimate

|(∂ sj f )(x)− (∂ s

j f )(x− e j

2m j)| ≤

∥∥∂ sj f

∥∥ .Λγ

(2|m j|)γ

and identity (3.3.9) to conclude the proof of (3.3.8).

The following is an immediate consequence.

Corollary 3.3.10. Let s ∈ Z with s≥ 0.

(a) Suppose that ∂α f exist and are integrable for all |α| ≤ s. Then for some constant

cn,s we have

| f (m)| ≤ cn,s

max(‖ f‖L1 ,max|α |=s ‖∂α f‖L1

)

(1+ |m|)s. (3.3.10)

(b) Suppose that ∂α f exist for all |α| ≤ s and whenever |α|= s, ∂α f are in.Λγ(T

n)for some 0 < γ < 1. Then for some constant c′n,s we have

| f (m)| ≤ c′n,smax

(‖ f‖L1 ,max|α |=s ‖∂α f‖Λγ

)

(1+ |m|)s+γ. (3.3.11)

Remark 3.3.11. The conclusions of Theorem 3.3.9 and Corollary 3.3.10 are also

valid when γ = 1. In this case the spaces Λγ should be replaced by the space Lip1

equipped with the seminorm

∥∥ f∥∥

Lip1= sup

x,h∈Tn

h =0

| f (x+h)− f (x)||h| .

There is a slight lack of uniformity in the notation here, since in the theory of Lips-

chitz spaces the notation.Λ1 is usually reserved for the space with seminorm

∥∥ f∥∥ .Λ1

= supx,h∈Tn

h =0

| f (x+h)+ f (x−h)−2 f (x)||h| .

The following proposition provides a partial converse to Theorem 3.3.9. We

denote below by [[s]] the largest integer strictly less than a given real number s.

Then [[s]] is equal to the integer part [s] of s, unless s is an integer, in which case

[[s]] = [s]−1.

Proposition 3.3.12. Let s > 0 and suppose that f is an integrable function on the

torus with

| f (m)| ≤C(1+ |m|)−s−n (3.3.12)

for all m ∈ Zn. Then f has partial derivatives of all orders |α| ≤ [[s]], and for 0 <γ < s− [[s]], ∂α f ∈

.Λγ for all multi-indices α satisfying |α|= [[s]].

Proof. Since f has an absolutely convergent Fourier series, Proposition 3.2.5 gives

that

f (x) = ∑m∈Zn

f (m)e2πix·m , (3.3.13)

for almost all x ∈ Tn.

Suppose that a series g=∑m gm satisfies∑m ‖∂ βgm‖L∞ <∞ for all |β | ≤M. Then

the function g is in C M and ∂ βg =∑m ∂βgm; indeed this can be proved by induction

on the degree of the multi-index, since for all |β | ≤M−1 we have

limt→0

∂ βg(x+ te j)−∂ βg(x)

t=∑

m

limt→0

∂ βgm(x+ te j)−∂ βgm(x)

t

=∑m

∂ j∂βg(x) ,

where the passage of the limit inside the sum is due to the Lebesgue dominated con-

vergence theorem, which can be applied using the uniform convergence of∑m ∂ j∂βg

via the mean value theorem.

Using the preceding observation, the function f in (3.3.13) is C [[s]](Tn) and

(∂α f )(x) = ∑m∈Zn

f (m)(2πim)αe2πix·m

for all multi-indices (α1, . . . ,αn) with |α| ≤ [[s]], since

∑m∈Zn

f (m) supx∈Tn

∣∣(2πim)αe2πix·m∣∣< ∞ ,

which holds because of (3.3.12).

Now suppose that |α|= [[s]] and that 0 < γ < s− [[s]]. Then

|(∂α f )(x+h)− (∂α f )(x)| =∣∣ ∑

m∈Zn

f (m)(2πim)αe2πix·m(e2πim·h−1)∣∣

≤ (2π)[[s]] ∑m∈Zn

|m|[[s]]C 21−γ(2π)γ |h|γ |m|γ(1+ |m|)n+s

≤C 21−γ(2π)s|h|γ ,

where we used the relation [[s]]+ γ− s−n <−n to conclude the convergence of the

series and the fact that

|e2πim·h−1| ≤min(2,2π|m| |h|)≤ 21−γ(2π)γ |m|γ |h|γ .

Next we recall the definition of functions of bounded variation.

Definition 3.3.13. A measurable function f on T1 is said to be of bounded variation

if it is defined everywhere and

Var( f ) = sup M

∑j=1

| f (x j)− f (x j−1)| : 0 = x0 < x1 < · · ·< xM = 1< ∞ ,

where the supremum is taken over all partitions of the interval [0,1]. The expression

Var( f ) is called the total variation of f . The class of functions of bounded variation

on T1 is denoted by BV (T1).

Examples of functions of bounded variation can be constructed as follows: given

f1, f2 nonnegative integrable functions on [0,1] with

∫ 1

0f1(t)dt =

∫ 1

0f2(t)dt ,

then the periodic function

g(x) =∫ x

0f1(t)dt−

∫ x

0f2(t)dt ,

defined on[0,1], is of bounded variation. Analogous examples can be constructed

when f1 and f2 are replaced by nonnegative finite Borel measures on [0,1].Every function of bounded variation can be represented as the difference of two

(not necessarily strictly) increasing functions and thus it has a finite derivative at

almost every point. Moreover, for functions of bounded variation, the Lebesgue–

Stieltjes integral with respect to d f is well defined.

Proposition 3.3.14. If f is in BV (T1), then

| f (m)| ≤ Var( f )

2π|m|

whenever m = 0.

Proof. Integration by parts gives

f (m) =∫

T1f (x)e−2πimx dx =

∫

T1

e−2πimx

−2πimd f ,

where the boundary terms vanish because of periodicity. The conclusion follows

from the fact that the norm of the measure d f is the total variation of f .

The following chart (Table 3.1) summarizes the decay of Fourier coefficients in

terms of scales of spaces measuring the smoothness of the functions. Recall that

for q ≥ 0, f (m) = o(|m|−q) means that | f (m)| |m|q → 0 as |m| → ∞ and f (m) =

O(|m|−q) means that | f (m)| ≤C |m|−q when |m| is large. In this chart, we denote by

C s,γ(Tn) the space of all C s functions on Tn, all of whose derivatives of total order

s lie in Λγ(Tn), for some 0 < γ < 1.

SPACE SEQUENCE OF FOURIER COEFFICIENTS

L1(Tn) o(1)

Lp(Tn) ℓp′ (Zn)

L2(Tn) ℓ2(Zn).Λγ (T

n) O(|m|−γ )BV (T1) O(|m|−1)

C 1(Tn) o(|m|−1)

C 1,γ (Tn) O(|m|−1−γ )C 2(Tn) o(|m|−2)

C 2,γ (Tn) O(|m|−2−γ )C 3(Tn) o(|m|−3)· · · · · ·

C ∞(Tn) o(|m|−N) for all N > 0

Table 3.1 Interconnection between smoothness of functions and decay of their Fourier coeffi-

cients. We take 0 < γ < 1 and 1 < p < 2.

3.3.3 Functions with Absolutely Summable Fourier Coefficients

Decay for the Fourier coefficients can also be indirectly deduced from knowledge

about the summability of these coefficients. The simplest kind of summability is in

the sense of ℓ1. It is therefore natural to consider the class of functions on the torus

whose Fourier coefficients form an absolutely summable series.

Definition 3.3.15. An integrable function f on the torus is said to have an absolutely

convergent Fourier series if

∑m∈Zn

| f (m)|<+∞.

We denote by A(Tn) the space of all integrable functions on the torus Tn whose

Fourier series are absolutely convergent. We then introduce a norm on A(Tn) by

setting ∥∥ f∥∥

A(Tn)= ∑

m∈Zn

| f (m)| .

In view of Proposition 3.2.5, every function f in A(Tn) can be changed on a set of

measure zero to be made continuous and under this modification, Fourier inversion

f (x) = ∑m∈Zn

f (m)e2πim·x

holds for all x ∈ Tn. Thus functions in A(Tn) are continuous and bounded. More-

over, Theorem 3.3.9 yields that every function in C n(Tn) whose partial derivatives

of order n are in.Λγ , γ > 0, must lie in A(Tn). The following theorem gives us a

significantly better sufficient condition for a function to be in A(Tn).

Theorem 3.3.16. Suppose f is a given function in C [n/2](Tn) and that all partial

derivatives of order [ n2] of f lie in

.Λγ(T

n) for some γ with n2− [ n

2] < γ < 1. Then f

lies in A(Tn) and

∥∥ f∥∥

A(Tn)≤ | f (0)|+C(n,γ) sup

|α |=[ n2 ]

∥∥∂α f∥∥ .Λγ (Tn)

,

where C(n,γ) is a constant depending on n and γ .

Proof. For each ℓ= 0,1,2, . . . , let

Sℓ =

(∑

2ℓ≤|m|<2ℓ+1

| f (m)|2)1/2

.

We begin by writing

∥∥ f∥∥

A(Tn)= | f (0)|+

∞

∑ℓ=0

∑2ℓ≤|m|<2ℓ+1

| f (m)| ≤ | f (0)|+√cn

∞

∑ℓ=0

2ℓn2 Sℓ , (3.3.14)

where we used the Cauchy-Schwarz inequality and the fact that there are at most

cn2ℓn points in Zn inside the open ball B(0,2ℓ+1), for some dimensional constant cn.

Notice that for a multi-index m = (m1, . . . ,mn) satisfying 2ℓ ≤ |m| ≤ 2ℓ+1 and for

j in 1, . . . ,n such that |m j|= supk |mk| we have

|m j|2ℓ

≥ |m|2ℓ√

n≥ 1√

n. (3.3.15)

For 1 ≤ j ≤ n, let e j be the element of Rn with zero entries except for the jth

coordinate, which is 1, and define

hℓj = 2−ℓ−2e j . (3.3.16)

Using the elementary fact that |t| ≤ π =⇒ |eit −1| ≥ 2|t|/π , we obtain

|e2πim·hℓj −1|= |e2πim j2−ℓ−2 −1| ≥ 2

π

|2πm j|2ℓ+2

=|m j|2ℓ

≥ 1√n, (3.3.17)

whenever|2πm j |2ℓ+2 ≤ π , which is always true since

|2πm j |2ℓ+2 ≤ 2π2ℓ+1

2ℓ+2 ≤ π .

We now have

S2ℓ =

n

∑j=1

∑2ℓ≤|m|<2ℓ+1

|m j |=supk |mk|

| f (m)|2

≤ nn

∑j=1

∑2ℓ≤|m|<2ℓ+1

|m j |=supk |mk|

|e2πim·hℓj −1|2| f (m)|2 |2πim j|2[n2 ]

|2πm j|2[n2 ]

≤ nn[

n2 ]

(2π2ℓ)2[ n2 ]

n

∑j=1∑

m∈Zn

|e2πim·hℓj −1|2|∂ [n/2]j f (m)|2

=Cn2−2ℓ[ n2 ]

n

∑j=1

∥∥∂ [n/2]j f ( · +hℓj)−∂

[n/2]j f

∥∥2

L2

≤Cn2−2ℓ[ n2 ]

n

∑j=1

∥∥∂ [n/2]j f ( · +hℓj)−∂

[n/2]j f

∥∥2

L∞

≤C′n2−2ℓ[ n2 ] sup|α |=[ n

2 ]

∥∥∂α f∥∥2.Λγ

n

∑j=1

|hℓj|2γ

=C′n,γ 2−2ℓ[ n2 ]−2ℓγ sup

|α |=[ n2 ]

∥∥∂α f∥∥2.Λγ

,

where we used (3.3.17), (3.3.15), and (3.3.16). We conclude that

Sℓ ≤C′′n,γ 2−ℓ([n2 ]+γ) sup

|α |=[ n2 ]

∥∥∂α f∥∥ .Λγ

which inserted in (3.3.14) yields the desired conclusion since γ > n2− [ n

2].

Exercises

3.3.1. Given a sequence an∞n=0 of positive numbers such that an → 0 as n→ ∞,

find a nonnegative integrable function h on [0,1] such that

∫ 1

0h(t)tm dt ≥ am.

Use this result to deduce a different proof of Lemma 3.3.2.[Hint: Try h = e

∞

∑k=0

(supj≥k

a j− supj≥k+1

a j)(k+2)χ[ k+1k+2

,1].]

3.3.2. Prove that given a positive sequence dmm∈Zn with dm → 0 as |m| → ∞,

there exists a positive sequence a j j∈Z with am1· · ·amn ≥ d(m1,...,mn) and a j → 0 as

| j| → ∞.

3.3.3. (a) Use the idea of the proof of Lemma 3.3.3 to prove that if a twice contin-

uously differentiable function f ≥ 0 is defined on (0,∞) and satisfies f ′(x)≤ 0 and

f ′′(x)≥ 0 for all x > 0, then limx→∞ x f ′(x) = 0.(b) Suppose that a continuously differentiable function g is defined on (0,∞) and

satisfies g≥ 0, g′ ≤ 0, and∫ ∞

1 g(x)dx <+∞. Prove that

limx→∞

xg(x) = 0.

3.3.4. Prove that for 0 < γ < δ < 1 we have ‖ f‖ .Λγ≤Cn,γ ,δ‖ f‖ .

Λδfor all functions

f and thus.Λδ is a subspace of

.Λγ .

3.3.5. Suppose that f is a differentiable function on T1 whose derivative f ′ is in

L2(T1). Prove that f ∈ A(T1) and that

∥∥ f∥∥

A(T1)≤

∥∥ f∥∥

L1 +1

2π

(∑j =0

j−2)1/2∥∥ f ′

∥∥L2 .

3.3.6. (a) Prove that the product of two functions in A(Tn) is also in A(Tn) and that

∥∥ f g∥∥

A(Tn)≤

∥∥ f∥∥

A(Tn)

∥∥g∥∥

A(Tn).

(b) Prove that the convolution of two square integrable functions on Tn always gives

a function in A(Tn).

3.3.7. Fix 0 < α < 1 and define f on T1 by setting

f (x) =∞

∑k=0

2−αke2πi2kx.

Prove that the function f lies in.Λα(T

1). Conclude that there does not exist positive

β > α such that for all f in.Λα(T

1) we have supm∈Z |m|β | f (m)|< ∞.[Hint: For h = 0 pick N ∈ Z+ such that 2N |h|> 1≥ 2N−1|h|. To estimate the differ-

ence | f (x+h)− f (x)|, consider the cases k ≤ N and k ≥ N +1 in the sum.]

3.3.8. Use without proof that there exists a constant C > 0 such that

supt∈R

∣∣∣∣N

∑k=2

eik logkeikt

∣∣∣∣≤C√

N, N = 2,3,4, . . . ,

to prove that the function

g(x) =∞

∑k=2

eik logk

ke2πikx

is in.Λ1/2(T

1) but not in A(T1). Conclude that the restriction s > 1/2 in Theorem

3.3.16 is sharp.[Hint: Estimate the difference |g(x+h)−g(x)| using the summation by parts iden-

tity in Appendix F, taking sums of the sequence eik logke2πikx and differences of the

sequence e2πikh−1k

.]

3.3.9. Show that there exist sequences amm∈Zn that tend to zero as |m| → ∞ for

which there do not exist functions f in L1(Tn) with f (m) = am for all m.[Hint: Suppose the contrary. Then the open mapping theorem would imply the in-

equality ‖ f‖L1(Tn) ≤ A‖ f ‖ℓ∞(Zn) for some A > 0. To contradict it, fix a smooth

nonzero function h equal to 1 on B(0, 14) and supported in B(0, 1

2). For b > 0 de-

fine gb(x) = h(x)e−π(1+ib)|x|2 and extend gb to a 1-periodic function in each variable

on Rn. Use that gb(m) =∫

Rn h(y)(1+ ib)−n/2e−π

1+ib|m−y|2dy, and let b → ∞ in the

inequality ‖gb‖L1(Tn) ≤ A‖gb ‖ℓ∞(Zn) to obtain a contradiction.]

3.4 Pointwise Convergence of Fourier Series

In this section we are concerned with the pointwise convergence of the square partial

sums and the Fejer means of a function defined on the torus.

3.4.1 Pointwise Convergence of the Fejer Means

We saw in Section 3.1 that the Fejer kernel is an approximate identity. This implies

that the Fejer (or Cesaro) means of an Lp function f on Tn converge to it in Lp for any

1≤ p <∞. Moreover, if f is continuous at x0, then the means (FnN ∗ f )(x0) converge

to f (x0) as N → ∞ in view of Theorem 1.2.19 (2). Although this is a satisfactory

result, it is natural to ask what happens for more general functions.

3.4 Pointwise Convergence of Fourier Series 205

Using properties of the Fejer kernel, we obtain the following one-dimensional

result regarding the convergence of the Fejer means:

Theorem 3.4.1. (Fejer) If a function f in L1(T1) has left and right limits at a point

x0, denoted by f (x0−) and f (x0+), respectively, then

(FN ∗ f )(x0)→1

2

(f (x0+)+ f (x0−)

)as N → ∞ . (3.4.1)

In particular, this is the case for functions of bounded variation.

Proof. Let us identify T1 with [−1/2,1/2]. Given ε > 0, find δ ∈ (0,1/2) such that

0 < t < δ =⇒∣∣∣∣

f (x0 + t)+ f (x0− t)

2− f (x0+)+ f (x0−)

2

∣∣∣∣< ε . (3.4.2)

Using the second expression for FN in (3.1.18), we can find an N0 > 0 such that for

N ≥ N0 we have

supδ≤t≤ 1

2

FN(t) =1

N +1sup

δ≤t≤ 12

(sin(π(N +1)t)

sin(πt)

)2

≤ 1

N +1

1

sin2(πδ )< ε . (3.4.3)

We now have

(FN ∗ f )(x0)− f (x0+) =∫

T1FN(t)

(f (x0 + t)− f (x0+)

)dt ,

(FN ∗ f )(x0)− f (x0−) =

∫

T1FN(t)

(f (x0− t)− f (x0−)

)dt .

Averaging these two identities and using that the integrand is even, we obtain

(FN ∗ f )(x0)−f (x0+)+ f (x0−)

2

= 2

∫ 1/2

0FN(t)

(f (x0 + t)+ f (x0− t)

2− f (x0+)+ f (x0−)

2

)dt .

(3.4.4)

We split the integral in (3.4.4) into two pieces, the integral over [0,δ ) and the inte-

gral over [δ ,1/2]. By (3.4.2), the integral over [0,δ ) is controlled by ε∫

T1 FN(t)dt =ε . Also (3.4.3) gives that for N ≥ N0

∣∣∣∣∫ 1/2

δFN(t)

(f (x0− t)+ f (x0 + t)

2− f (x0−)+ f (x0+)

2

)dt

∣∣∣∣

≤ ε

2

(∥∥ f − f (x0−)∥∥

L1 +∥∥ f − f (x0+)

∥∥L1

)= ε c( f ,x0) ,

where c( f ,x0) is a constant depending on f and x0. We have now proved that given

ε > 0 there exists an N0 such that for N ≥ N0 the second expression in (3.4.4) is

bounded by 2ε (c( f ,x0)+1). This proves the required conclusion.

Functions of bounded variation can be written as differences of increasing func-

tions, and since increasing functions have left and right limits everywhere, (3.4.1)

holds for these functions.

We continue with an elementary but very useful application of the preceding

result.

Proposition 3.4.2. Let x0 ∈ T1 and let f be a complex-valued function on T1. Sup-

pose that the left and right limits of f exist as x → x0 and that the partial sums

(Dirichlet means) (DN ∗ f )(x0) converge. Then

(DN ∗ f )(x0)→1

2

(f (x0+)+ f (x0−)

)

as N → ∞.

Proof. If (DN ∗ f )(x0)→ L(x0) as N → ∞, then

(FN ∗ f )(x0) =(D0 ∗ f )(x0)+(D1 ∗ f )(x0)+ · · ·+(DN ∗ f )(x0)

N +1→ L(x0)

as N → ∞. But (FN ∗ f )(x0)→ 12

(f (x0+)+ f (x0−)

)as N → ∞ in view of Theorem

3.4.1. We conclude that

L(x0) =1

2

(f (x0+)+ f (x0−)

),

Thus (DN ∗ f )(x0)→ 12

(f (x0+)+ f (x0−)

)as N → ∞.

This theorem is quite useful when we have a priori knowledge that the Fourier

series converges. For instance, consider the following example.

Example 3.4.3. On (−1/2,1/2) let f (t) = t and f (1/2) = f (−1/2) = 1000. Then

f is discontinuous at the point −1/2 ≡ 1/2 but it has left and right limits at this

point:

limt→− 1

2+f (t) =−1

2lim

t→ 12−

f (t) =1

2. (3.4.5)

Moreover f (m) = i(−1)m

2πmwhen m = 0 and f (0) = 0 by Exercise 3.4.1 (a). It is not

hard to see that the series

(DN ∗ f )(x) =i

2π ∑0<|m|≤N

(−1)m

me2πimx =

i

2π ∑0<|m|≤N

e2πim(x+ 12 )

m(3.4.6)

converges for every x ∈ (−1/2,1/2). Indeed, by Appendix F, (3.4.6) equals

i

2π

1

N

N

∑m=1

(e2πim(x+ 1

2 )− e−2πim(x+ 12 ))

− i

2π

N−1

∑k=1

(k

∑m=1

(e2πim(x+ 1

2 )− e−2πim(x+ 12 )))(

1

k+1− 1

k

)

which has a limit as N → ∞, since the geometric sums

N

∑m=1

e±2πim(x+ 12 ) =

1− e±2πi(N+1)(x+ 12 )

1− e±2πi(x+ 12 )

−1

are bounded above independently of N when x ∈ (−1/2,1/2). We conclude that

f (x) = x = limN→∞

i

2π ∑0<|m|≤N

e2πim(x+ 12 )

m=− lim

N→∞∑

0<|m|≤N

sin(2πm(x+ 12))

2πm

whenever |x|< 1/2. Moreover, we have that

(DN ∗ f )(1/2) = limN→∞

i

2π ∑0<|m|≤N

0

m= 0 ,

which is the average of the left and right limits in (3.4.5) as Proposition 3.4.2 states.

Exercise 3.4.2 contains other applications of this sort.

3.4.2 Almost Everywhere Convergence of the Fejer Means

We have seen that the Fejer means of a relatively nice function (such as of bounded

variation) converge everywhere. What can we say about the Fejer means of a general

integrable function? Since the Fejer kernel is an approximate identity that satisfies

good estimates, the following result should not come as a surprise.

Theorem 3.4.4. (a) For f ∈ L1(Tn), let

H ( f ) = supN∈Z+

| f ∗FnN | .

Then H maps L1(Tn) to L1,∞(Tn) and Lp(Tn) to itself for 1 < p≤ ∞.

(b) For any function f ∈ L1(Tn), we have as N → ∞

(FnN ∗ f )→ f a.e.

Proof. It is an elementary fact that |t| ≤ π2

=⇒ |sin t| ≥ 2π |t|; see Appendix E.

Using this fact and the expression (3.1.18) we obtain for all t in [− 12, 1

2],


|FN(t)| =1

N +1

∣∣∣∣sin(π(N +1)t)

sin(πt)

∣∣∣∣2

≤ N +1

4

∣∣∣∣sin(π(N +1)t)

(N +1)t

∣∣∣∣2

≤ N +1

4min

(π2,

1

(N +1)2t2

)

≤ π2

2

N +1

1+(N +1)2|t|2 .

For t ∈ R let us set ϕ(t) = (1 + |t|2)−1 and ϕε(t) =1ε ϕ(

tε ) for ε > 0. For x =

(x1, . . . ,xn) ∈ Rn and ε > 0 we also set

Φ(x) = ϕ(x1) · · ·ϕ(xn)

and Φε(x) = ε−nΦ(ε−1x). Then for |t| ≤ 12

we have |FN(t)| ≤ π2

2ϕε(t) with ε =

(N +1)−1, and for y ∈ [− 12, 1

2]n we have

|FnN(y)| ≤ (π

2

2)nΦε(y), with ε = (N +1)−1.

Now let f be an integrable function on Tn and let f0 denote its periodic extension

on Rn. For x ∈ [− 12, 1

2]n we have

H ( f )(x) = supN>0

∣∣∣∣∫

TnFn

N(y) f (x− y)dy

∣∣∣∣

≤ (π2

2)n sup

ε>0

∫

[− 12 ,

12 ]

n|Φε(y)| | f0(x− y)|dy

≤ 5n supε>0

∫

Rn|Φε(y)| |( f0χQ)(x− y)|dy

= 5nG ( f0χQ)(x),

(3.4.7)

where Q is the cube [−1,1]n and G is the operator defined on integrable functions

on Rn by

G (h) = supε>0

|h| ∗Φε .

If we can show that G maps L1(Rn) to L1,∞(Rn), the corresponding conclusion for

H on Tn would follow from the fact H ( f )≤ 5nG ( f0χQ) proved in (3.4.7) and the

sequence of inequalities

∥∥H ( f )∥∥

L1,∞(Tn)≤ 5n

∥∥G ( f0χQ)∥∥

L1,∞(Rn)≤ 5nC

∥∥ f0χQ

∥∥L1(Rn)

= C′∥∥ f

∥∥L1(Tn)

.


Moreover, the Lp conclusion about H follows from the weak type (1,1) result and

the trivial L∞ inequality, in view of the Marcinkiewicz interpolation theorem (The-

orem 1.3.2). The required weak type (1,1) estimate for G on Rn is a consequence of

Lemma 3.4.5. Modulo the proof of this lemma, part (a) of the theorem is proved.

To prove the statement in part (b) observe that for f ∈ C ∞(Tn), which is a dense

subspace of L1, we have FnN ∗ f → f uniformly on Tn as N → ∞, since the sequence

FNN is an approximate identity. Since by part (a), H maps L1(Tn) to L1,∞(Tn),Theorem 2.1.14 yields that for f ∈ L1(Tn), Fn

N ∗ f → f a.e.

We now prove the weak type (1,1) boundedness of G used earlier.

Lemma 3.4.5. Let Φ(x1, . . . ,xn) = (1+ |x1|2)−1 · · ·(1+ |xn|2)−1 and for ε > 0 let

Φε(x) = ε−nΦ(ε−1x). Then the maximal operator

G ( f ) = supε>0

| f | ∗Φε

maps L1(Rn) to L1,∞(Rn).

Proof. Let I0 = [−1,1] and Ik = t ∈ R : 2k−1 ≤ |t| ≤ 2k for k = 1,2, . . . . Also, let

Ik be the convex hull of Ik, that is, the interval [−2k,2k]. For a2, . . . ,an fixed positive

numbers, let Ma2,...,an be the maximal operator obtained by averaging a function on

Rn over all products of closed intervals J1×·· ·× Jn containing a given point with

|J1|= 2a2 |J2|= · · ·= 2an |Jn|.

In view of Exercise 2.1.9(c), we have that Ma2,...,an maps L1 to L1,∞ with some con-

stant independent of the a j’s. (This is due to the nice doubling property of this family

of rectangles.) For a fixed ε > 0 we estimate the expression

(Φε ∗ | f |)(0) =∫

Rn

| f (−εy)|dy

(1+ y21) · · ·(1+ y2

n).

Split Rn into n! regions of the form |y j1 | ≥ · · · ≥ |y jn |, where j1, . . . , jn is a per-

mutation of the set 1, . . . ,n and y = (y1, . . . ,yn). By symmetry, we examine the

region R where |y1| ≥ · · · ≥ |yn|. Then for some constant C > 0 we have

∫

R

| f (−εy)|dy

(1+y21) · · ·(1+y2

n)≤C

∞

∑k1=0

k1

∑k2=0

· · ·kn−1

∑kn=0

2−(2k1+···+2kn)∫

Ik1

· · ·∫

Ikn

| f (−εy)|dykn· · ·dy1,

and the last expression is trivially controlled by the corresponding expression, where

the Ik’s are replaced by the Ik’s. This, in turn, is controlled by

C′∞

∑k1=0

k1

∑k2=0

· · ·kn−1

∑kn=0

2−(k1+···+kn)Mk1−k2,...,k1−kn( f )(0) . (3.4.8)


Now set s2 = k1− k2, . . . ,sn = k1− kn, observe that s j ≥ 0, use that

2−(k1+···+kn) ≤ 2−k12 2−

s22n · · ·2− sn

2n ,

and change the indices of summation to estimate the expression in (3.4.8) by

C′′∞

∑k1=0

∞

∑s2=0

· · ·∞

∑sn=0

2−k12 2−

s22n · · ·2− sn

2n Ms2,...,sn( f )(0) .

Argue similarly for the remaining regions |y j1 | ≥ · · · ≥ |y jn |. Finally, translate to an

arbitrary point x to obtain the estimate

|(Φε ∗ f )(x)| ≤C′′n!∞

∑s2=0

· · ·∞

∑sn=0

2−s22n · · ·2− sn

2n Ms2,...,sn( f )(x) .

Now take the supremum over all ε > 0 and use the fact that the maximal functions

Ms2,...,sn map L1 to L1,∞ uniformly in s2, . . . ,sn as well as the result of Exercise 1.4.10

to obtain the desired conclusion for G .

3.4.3 Pointwise Divergence of the Dirichlet Means

We now pass to the more difficult question of convergence of the square partial sums

of a Fourier series. It is natural to start our investigation with the class of continuous

functions. Do the partial sums of the Fourier series of continuous functions converge

pointwise? The following simple proposition warns about the behavior of partial

sums.

Proposition 3.4.6. (a) (duBois Reymond) There exists a continuous function f on

T1 whose partial sums diverge at a point. Precisely, for some point x0 ∈ T1 we have

limsupN→∞

∣∣∣∣ ∑m∈Z|m j |≤N

f (m)e2πix0m

∣∣∣∣= ∞ .

(b) There exists a continuous function F on Tn and x0 ∈ T1 such that the sequence

limsupN→∞

∣∣∣∣ ∑m∈Zn

|m j |≤N

F(m)e2πi(x0m1+x2m2+···+xnmn)

∣∣∣∣= ∞

for all x2, . . . ,xm in T1.


Proof. The proof of part (b) is obtained by considering the continuous function

F(x1, . . . ,xn) = f (x1), where f is as in part (a). Then we have

(F ∗DnN)(x1, . . . ,xn) = ( f ∗DN)(x1)

and thus the square partial sums of F diverge on the (n− 1)-dimensional plane

(x0,x2, . . . ,xn) : x2, . . . ,xn ∈ T1.We now prove part (a) using functional analysis. For a constructive proof, see

Exercise 3.4.7. Let C(T1) be the Banach space of all continuous functions on the

circle equipped with the L∞ norm. Consider the continuous linear functionals

f → TN( f ) = (DN ∗ f )(0)

on C(T1) for N = 1,2, . . . . We show that the norms of the TN’s on C(T1) converge to

infinity as N →∞. To see this, given any integer N ≥ 100, let ϕN(x) be a continuous

even function on [− 12, 1

2] that is bounded by 1 and is equal to the sign of DN(x)

except at small intervals of length (2N +1)−2 around the 2N +1 zeros of DN . Call

the union of all these intervals BN and set AN = [− 12, 1

2]\BN . Then

∫

BN

∣∣DN(x)∣∣dx+

∣∣∣∣∫

BN

ϕN(x)DN(x)dx

∣∣∣∣≤ 2 |BN |(2N +1) = 2.

Using this estimate we obtain

∥∥TN

∥∥C(T1)→C

≥ |TN(ϕN)|=∣∣∣∣∫

T1DN(−x)ϕN(x)dx

∣∣∣∣

≥∫

AN

∣∣DN(x)∣∣dx−

∣∣∣∣∫

BN

DN(x)ϕN(x)dx

∣∣∣∣

=∫

T1

∣∣DN(x)∣∣dx−

∣∣∣∣∫

BN

DN(x)ϕN(x)dx

∣∣∣∣−∫

BN

∣∣DN(x)∣∣dx

≥ 4

π2

N

∑k=1

1

k−2 .

It follows that the norms of the linear functionals TN are not uniformly bounded. The

uniform boundedness principle now implies the existence of a function f ∈C(T1)and of a sequence N j → ∞ such that

|TN j( f )| → ∞

as j→ ∞. The Fourier series of this f diverges at x0 = 0.

3.4.4 Pointwise Convergence of the Dirichlet Means

We have seen that continuous functions may have divergent Fourier series. How

about Lipschitz continuous functions? As it turns out, there is a more general condi-

tion that implies convergence for the Fourier series of functions that satisfy a certain

integrability condition.

Theorem 3.4.7. (Dini) Let f be an integrable function on T1, let t0 be a point on T1

for which f (t0) is defined and assume that

∫

|t|≤ 12

| f (t + t0)− f (t0)||t| dt < ∞ . (3.4.9)

Then (DnN ∗ f )(t0)→ f (t0) as N → ∞.

(Tonelli) Let f be an integrable function on Tn and let a = (a1, . . . ,an) ∈ Tn. If f is

defined at a and

∫

|x1|≤ 12

· · ·∫

|xn|≤ 12

| f (x+a)− f (a)||x1| · · · |xn|

dxn · · ·dx1 < ∞ , (3.4.10)

then we have (DnN ∗ f )(a)→ f (a) as N → ∞.

Proof. Since the one-dimensional result is contained in the multidimensional one,

we prove the latter. Replacing f (x) by f (x+ a)− f (a), we may assume that a = 0

and f (a) = 0. Using identities (3.1.15) and (3.1.14), we can write

(DnN ∗ f )(0) =

∫

Tnf (−x)

n

∏j=1

sin((2N +1)πx j)

sin(πx j)dxn · · ·dx1 (3.4.11)

=∫

Tnf (−x)

n

∏j=1

(sin(2Nπx j)cos(πx j)

sin(πx j)+ cos(2Nπx j)

)dxn · · ·dx1 .

Expand out the product to express the integrand as a sum of terms of the form

f (−x)∏

j∈I

cos(πx j)

sin(πx j)

∏j∈I

sin(2Nπx j) ∏k∈1,2,...,n\I

cos(2Nπxk) , (3.4.12)

where I is a subset of 1,2, . . . ,n; here we use the convention that the product over

an empty set of indices is 1. The function fI inside the curly brackets in (3.4.12)

is integrable on [− 12, 1

2)n except possibly in a neighborhood of the origin, since

|sin(πx j)| ≥ 2|x j| when |x j| ≤ 12. But condition (3.4.10) with a = 0 and f (a) = 0

guarantees that fI is also integrable in a neighborhood of the origin. Expressing the

sines and cosines in (3.4.12) in terms of exponentials, we obtain that the integral of

(3.4.12) over [− 12, 1

2)n is a finite linear combination of Fourier coefficients of fI at

the points (±N, . . . ,±N) ∈ Zn. Applying Lemma 3.3.1 yields that the expression in

(3.4.11) tends to zero as N → ∞.

The following are consequences of this test.

Corollary 3.4.8. (a) (Riemann’s principle of localization) Let f be an integrable

function on T1 that vanishes on an open interval I. Then DN ∗ f converges to zero

on the interval I.

(b) Let a = (a1, . . . ,an) ∈ Tn and suppose that an integrable function f on Tn is

constant on the cross

x = (x1, . . . ,xn) ∈ Tn : |x j−a j|< δ j for some j ,

where 0 < δ j < 1/2 are fixed. Then (DnN ∗ f )(a)→ f (a) as N → ∞.

Proof. (a) Let t0 ∈ I. If f vanishes on I, condition (3.4.9) holds, since the function

t → f (t + t0)− f (t0) vanishes on −t0 + I, which is an interval containing the origin,

and is integrable outside −t0 + I. Thus (DN ∗ f )(t0)→ f (t0) = 0 for every t0 ∈ I.

(b) We need to show that the function

| f (x+a)− f (a)||x1| · · · |xn|

is integrable over Tn = [−1/2,1/2)n. The integral of this function over Tn is equal

to its integral over the region

S = (x1, . . . ,xn) ∈ Tn : |xk| ≥ δk for all k ,

since f (x+a)− f (a) vanishes whenever |x j|< δ j for some j ∈ 1,2, . . . ,n. But on

S we have that| f (x+a)− f (a)||x1| · · · |xn|

≤ | f (x+a)− f (a)|δ1 · · ·δn

and this function is integrable over S, since f is. We deduce that (3.4.10) holds.

Corollary 3.4.9. Let a ∈ Tn and suppose that f ∈ L1(Tn) satisfies

| f (x)− f (a)| ≤C|x1−a1|ε1 · · · |xn−an|εn

for some C,ε j > 0 and for all x ∈ Tn. Then the square partial sums (DnN ∗ f )(a)

converge to f (a).

Proof. Note that condition (3.4.10) holds.

Corollary 3.4.10. (Dirichlet) If f is defined on T1 and is a differentiable function

at a point a in T1, then (DN ∗ f )(a)→ f (a).

Proof. There exists a δ > 0 (say less than 1/2) such that | f (x)− f (a)|/|x− a| is

bounded by | f ′(a)|+ 1 for |x− a| ≤ δ . Also | f (x)− f (a)|/|x− a| is bounded by

| f (x)− f (a)|/δ when |x−a|> δ . It follows that condition (3.4.9) holds.

Exercises

3.4.1. Identify T1 with [−1/2,1/2) and fix 0 < b < 1/2. Prove the following:

(a) The mth Fourier coefficient of the function x is i(−1)m

2πmwhen m = 0 and 0 when

m = 0.

(b) The mth Fourier coefficient of the function χ[−b,b] issin(2πbm)

mπ when m = 0 and

2b when m = 0.

(c) The mth Fourier coefficient of the function(1− |x|

b

)+

issin2(πbm)

bm2π2 when m = 0

and b when m = 0.

(d) The mth Fourier coefficient of the function |x| is − 12m2π2 +

(−1)m

2m2π2 when m = 0

and 14

when m = 0.

(e) The mth Fourier coefficient of the function x2 is(−1)m

2m2π2 when m = 0 and 112

when m = 0.

(f) The mth Fourier coefficient of the function cosh(2πx) is(−1)m

1+m2sinhππ .

(g) The mth Fourier coefficient of the function sinh(2πx) isim(−1)m

1+m2sinhππ .

3.4.2. Use Exercise 3.4.1 and Proposition 3.4.2 to prove that

∑k∈Z

1

(2k+1)2=π2

4∑

k∈Z\0

1

k2=π2

3

∑k∈Z\0

(−1)k+1

k2=π2

6∑k∈Z

(−1)k

k2 +1=

2π

eπ − e−π.

3.4.3. Let M > N be given positive integers.

(a) For f ∈ L1(T1), prove the following identity:

(DN ∗ f )(x) =M+1

M−N(FM ∗ f )(x)− N +1

M−N(FN ∗ f )(x)

− M+1

M−N∑

N<| j|≤M

(1− | j|

M+1

)f ( j)e2πi jx .

(b) (G. H. Hardy) Suppose that a function f on T1 satisfies the following condition:

for any ε > 0 there exists an a > 1 and a k0 > 0 such that for all k ≥ k0 we have

∑k<|m|≤[ak]

| f (m)|< ε .

Use part (a) to prove that if (FN ∗ f )(x) converges (uniformly) to A(x) as N → ∞,

then (DN ∗ f )(x) also converges (uniformly) to A(x) as N → ∞.

3.4.4. Use Proposition 3.4.2 to show that for 0 < b < 12

we have

limN→∞

N

∑m=−N

m=0

sin(2πbm)

mπe2πibm =

1

2−2b .

[Hint: Use Exercise 3.4.1(b).

]

3.4.5. Let f be an integrable function on Tn and g be a bounded function on Tn

and let K be a compact subset of Tn. Consider the family F = fw : w ∈ K, where

fw(x) = f (x−w)g(x) for all x∈Tn. Prove that the Riemann–Lebesgue lemma holds

uniformly for the family F . This means that given ε > 0 there exists an N0(K)> 0

such that for |m| ≥ N0 we have | fw(m)| ≤ ε for all w ∈ K.

3.4.6. Prove the following version of Corollary 3.4.8 (b). Suppose that a function f

on Tn is constant on the cross U = (x1, . . . ,xn) ∈ Tn : |x j− a j| < δ for some j,for some δ < 1/2. Then Dn

N ∗ f converges to f (a) uniformly on compact subsets of

the box W = (x1, . . . ,xn) ∈ Tn : |x j|< δ for all j.[Hint: Use Exercise 3.4.5.

]

3.4.7. Follow the steps given to obtain a constructive proof of the existence of a

continuous function whose Fourier series diverges at a point. Identify T1 with [0,1)and define

g(x) =−2πi(x−1/2).

(a) Prove that g(m) = 1/m when m = 0 and zero otherwise.

(b) Prove that for all nonnegative integers M and N we have

((e2πiN( ·)(g∗DN))∗DM

)(x) = e2πiNx ∑

1≤|r|≤N

1

re2πirx

when M ≥ 2N and

((e2πiN( ·)(g∗DN))∗DM

)(x) = e2πiNx ∑

−N≤r≤M−Nr =0

1

re2πirx

when M < 2N. Conclude that there exists a constant C > 0 such that for all M, N,

and x = 0 we have

|(e2πiN( ·)(g∗DN)

)∗DM(x)| ≤ C

|x| .

(c) Show that there exists a constant C1 > 0 such that

supN>0

supx∈T1

∣∣(g∗DN)(x)∣∣= sup

N>0

supx∈T1

∣∣∣∣∣ ∑1≤|r|≤N

1

re2πirx

∣∣∣∣∣≤C1 < ∞ .

(d) Let λk = 1+ eek. Define

f (x) =∞

∑k=1

1

k2e2πiλkx(g∗Dλk

)(x)

and prove that f is continuous on T1 and that its Fourier series converges at every

x = 0, but limsupM→∞ |( f ∗DM)(0)|= ∞.[Hint: Take M = eem

with m→ ∞. The inequality in part (b) follows by summation

by parts.]

3.5 A Tauberian theorem and Functions of Bounded Variation

The relation between the partial sums of a Fourier series and the Fejer means is a

particular situation of a relation between sequences of complex numbers and their

arithmetic means. Given a sequence ak∞k=0 of complex numbers, we denote its

partial sums by

sN = a1 + · · ·+aN

for N ≥ 0, and its arithmetic or Cesaro means by

σN =1

N +1

N

∑k=0

sk =1

N +1

N

∑k=0

(N +1− k)ak .

A classical result says that if sN → L as N→∞, then σN → L as N→∞. The converse

is not true, as the example ak = (−1)k indicates. But in a particular situation the

reverse implication holds.

3.5.1 A Tauberian theorem

We have the following result concerning the convergence of sk∞k=0 as a conse-

quence of that of σk∞k=0.

Theorem 3.5.1. (a) Suppose that for a sequence ak∞k=0 of complex numbers we

have that σN → L as N → ∞ and that |kak| ≤ M < ∞ for all k = 0,1,2, . . . . Then

sk → L as k→ ∞.

(b) Let X be a nonempty set. Suppose that for a sequence ak(x)∞k=0 of complex-

valued functions on X we have that σN(x)→ L(x) uniformly in x ∈ X as N → ∞and that supk≥0 supx∈X |kak(x)| ≤M < ∞. Then sk(x)→ L(x) uniformly in x ∈ X as

k→ ∞.

3.5 A Tauberian theorem and Functions of Bounded Variation 217

Proof. We prove part (b), noting that the proof of part (a) is subsumed in that of (b).

For 0≤ k < m < ∞ we have

(m+1)σm(x)− (k+1)σk(x)−m

∑j=k+1

(m+1− j)a j(x)

=m

∑j=0

(m+1− j)a j(x)−k

∑j=0

(k+1− j)a j(x)−m

∑j=k+1

(m+1− j)a j(x)

=(m− k)k

∑j=0

a j(x)

=(m− k)sk(x) .

Therefore we have

m+1

m− kσm(x)−

k+1

m− kσk(x)−

1

m− k

m

∑j=k+1

(m+1− j)a j(x) = sk(x)

and thus

sk(x)−σk(x) =m+1

m− k(σm(x)−σk(x))−

m+1

m− k

m

∑j=k+1

(1

j− 1

m+1

)ja j(x) . (3.5.1)

Notice that

m

∑j=k+1

(1

j− 1

m+1

)≤

∫ m

k

dt

t−

m

∑j=k+1

1

m+1= log

m

k− m− k

m+1. (3.5.2)

Now fix ε > 0 such that ε < 1. For each k ∈ Z+ pick an mk ∈ k,k+1, . . . ,2k such

thatmkk→ 1+ ε . Then

mk+1mk−k

converges to ε−1 +1 as k→ ∞, hence it is bounded by

some constant Cε . Then (3.5.1) and (3.5.2) with mk in place of m yield

supx∈X

∣∣sk(x)−σk(x)∣∣≤Cε sup

x∈X

|σmk(x)−σk(x)|+M

mk +1

mk− k

[log

mk

k− mk− k

mk +1

].

Taking the limsupk→∞ in the preceding inequality and using that

limsupk→∞

supx∈X

|σmk(x)−σk(x)|= 0 ,

which is a consequence of the hypothesis that σk (and thus σmk) converges to L

uniformly, we obtain

limsupk→∞

supx∈X

∣∣sk(x)−σk(x)∣∣≤M

[(1+

1

ε

)log(1+ ε)−1

].

In view of the Taylor expansion

log(1+ ε) = ε− 1

2ε2 +

1

3ε3−·· ·= ε+O(ε2) ,

which is valid for 0 < ε < 1, we conclude that

limsupk→∞

supx∈X

∣∣sk(x)−σk(x)∣∣≤M cε

for some absolute constant c > 0. Since ε > 0 was arbitrary, we finally deduce that

sk(x) converges uniformly to the same limit as σk(x), which is L(x).

Corollary 3.5.2. Suppose that a function f on T1 is continuous and there is a con-

stant M > 0 such that | f (m)| ≤M|m|−1 for all m∈Z+ \0. Then the Fourier series

of f converges uniformly to f . In particular, if f is a continuous function of bounded

variation on the circle, then f ∗DN → f uniformly on T1 as N → ∞.

Proof. The Fejer means FN∞N=0 are an approximate identity on Tn (Proposition

3.1.10) and so FN ∗ f converge uniformly to f on T1 as N → ∞ in view of Theorem

1.2.19 (2). Moreover, we have |m| | f (m)| ≤M for all m∈Z. It follows from Theorem

3.5.1 that DN ∗ f converges uniformly to f .

If, additionally, f is a function of bounded variation, then |m| | f (m)| ≤ 12π Var( f ),

as shown in Proposition 3.3.14. Then the claimed conclusion follows.

3.5.2 The sine integral function

We examine a few useful properties of the antiderivative of sin(t)/t.

Definition 3.5.3. For 0≤ x < ∞ define the sine integral function

Si(x) =∫ x

0

sin(t)

tdt . (3.5.3)

Integrating by parts we write

Si(x) =∫ 1

0

sin(t)

tdt +

−cos(x)

x+ cos(1)−

∫ x

1

cos(t)

t2dt ,

from which it follows that the limit of Si(x) as x→ ∞ exists and is equal to

limx→∞

Si(x) =∫ 1

0

sin(t)

tdt + cos(1)−

∫ ∞

1

cos(t)

t2dt .

To precisely evaluate the limit of Si(x) as x→ ∞ we write

Si((N + 12)π) =π

∫ 12

0

sin((2N +1)πt)

πtdt

=π∫ 1

2

0

sin((2N +1)πt)

sin(πt)dt +π

∫ 12

0sin((2N +1)πt)

1

πt− 1

sin(πt)

dt

=π

2

∫ 12

− 12

DN(t)dt +π

2

∫ 12

− 12

e(2N+1)πit − e−(2N+1)πit

2i

1

πt− 1

sin(πt)

dt,

which converges to π/2+ 0 as N → ∞, in view of the Riemann-Lebesgue lemma

(Proposition 3.3.1), since the function inside the curly brackets is integrable over the

circle. We conclude that limx→∞ Si(x) = π/2.

Note that Si′ vanishes at nπ , n = 0,1,2, . . . and Si′′(nπ) = (−1)n/nπ . Conse-

quently, Si(x) has local maxima at the points π , 3π , 5π , . . . and local minima at the

points 2π , 4π , 6π , . . . . Moreover, it is increasing on the intervals [2kπ,(2k+ 1)π]and decreasing on [(2k+1)π,(2k+2)π], k = 0,1,2, . . . . Also, observe that

Si(3π)−Si(π)=∫ 2π

π

sin(t)

tdt+

∫ 2π

π

sin(t +π)

t +πdt =

∫ 2π

πsin(t)

(1

t− 1

t +π

)dt < 0

and likewise we can prove the remaining inequalities in the sequence

Si(π)> Si(3π)> Si(5π)> Si(7π)> · · ·> π

2.

Similarly, one can show that

Si(2π)< Si(4π)< Si(6π)< · · ·< π

2.

Hence Si(π) is the absolute maximum of Si(x) on [0,∞), while 0 is the absolute

minimum of Si(x) on [0,∞); Si(π) is the absolute minimum of Si(x) on [π,∞).

3.5.3 Further properties of functions of bounded variation

Next we have the following theorem concerning functions of bounded variation.

Recall that functions of bounded variation are differences of increasing functions

and thus have left and right limits at every point.

Theorem 3.5.4. Let 0 < δ ≤ 1/2. Suppose that f is an integrable function on T1

which is of bounded variation on the neighborhood [t0 − δ , t0 + δ ] of the point

t0 ∈ T1. Then

limN→∞

( f ∗DN)(t0) =f (t0+)+ f (t0−)

2.


Proof. We write W for the neighborhood (−δ ,δ ) of 0, Ft0(t) =f (t0−t)+ f (t0+t)

2, and

Lt0 =f (t0+)+ f (t0−)

2. We have

( f ∗DN)(t0) =

∫

T1f (t0− t)DN(t)dt =

∫

T1f (t0 + t)DN(t)dt ,

hence, averaging yields

( f ∗DN)(t0) =∫

T1

f (t0− t)+ f (t0 + t)

2DN(t)dt =

∫

T1Ft0(t)DN(t)dt .

Therefore we have

( f ∗DN)(t0)−Lt0 =∫

W

(Ft0(t)−Lt0

)DN(t)dt +

∫

T1\W

(Ft0(t)−Lt0

)DN(t)dt

and since in the second integral |t| ≥ δ , the Riemann-Lebesgue lemma shows that

the second term is o(1), i.e., it tends to zero as N → ∞. We now show that the first

integral also goes to zero. We write

∫

W

(Ft0(t)−Lt0

)DN(t)dt =

∫

W

(Ft0(t)−Lt0

) sin((2N +1)πt)

πtdt (3.5.4)

+∫

W

(Ft0(t)−Lt0

)( 1

sin(πt)− 1

πt

)sin((2N +1)πt)dt ,

but since the function 1πt− 1

sin(πt) remains bounded on [− 12, 1

2] (Exercise 3.1.5 (a)), it

follows from the Riemann-Lebesgue lemma that the second term is o(1) as N → ∞.

Consequently,

( f ∗DN)(t0)−Lt0 =1

π

∫

W

(Ft0(t)−Lt0

) sin((2N +1)πt)

tdt +o(1)

as N → ∞. To prove the required conclusion, it will suffice to show that

2

π

∫ δ

0

(Ft0(t)−Lt0

) sin((2N +1)πt)

tdt → 0 (3.5.5)

as N → ∞. Let Si(t) be as defined in (3.5.3). We express the integral in (3.5.5) as

∫ δ

0

(Ft0(t)−Lt0

)Si′((2N +1)πt)dt (3.5.6)

Integrating by parts we obtain that (3.5.6) is equal to

(Ft0(δ−)−Lt0)Si((2N +1)πδ )−∫ δ

0Si((2N +1)πt)dFt0(t) . (3.5.7)


Letting N → ∞ and using the Lebesgue dominated convergence theorem, we con-

clude that (3.5.7) converges to

(Ft0(δ−)−Lt0)π

2−

∫ δ

0

π

2dFt0(t) = (Ft0(δ−)−Lt0)

π

2− (Ft0(δ−)−Ft0(0+))

π

2= 0

noticing that Lt0 = Ft0(0+).

Next, we obtain an explicit bound for the partial sums of functions of bounded

variation. Let Si(t) be as in (3.5.3).

Theorem 3.5.5. Suppose that f is a function of bounded variation on the circle T1.

Then the partial sums of the Fourier series of f are uniformly bounded, in particular,

we have

supt0∈T1

supN∈Z+

|( f ∗DN)(t0)| ≤(

1− 2

π+Si(π)

)‖ f‖L∞ +Si(π)Var( f ) . (3.5.8)

Proof. We take δ = 1/2 in the proof of the preceding theorem. For a point t0 ∈ T1,

let Ft0(t) =f (t0−t)+ f (t0+t)

2. We have that

( f ∗DN)(t0) =∫

T1Ft0(t)

sin((2N +1)πt)

πtdt

+

∫

T1Ft0(t)

(1

sin(πt)− 1

πt

)sin((2N +1)πt)dt .

(3.5.9)

Using that∣∣ 1

sin(πt) −1πt

∣∣≤ 1− 2π when |t| ≤ 1

2(Exercise 3.1.5 (a)), we obtain that the

second integral in (3.5.9) is bounded by (1− 2π )‖ f‖L∞ . Integrating by parts as in the

proof of the preceding theorem, we express the first integral in (3.5.9) as

Ft0(12−)Si((2N +1)π 1

2)−

∫ 12

0Si((2N +1)πt)dFt0(t) , (3.5.10)

which is bounded (in absolute value) by ‖ f‖L∞Si(π) + Si(π)Var( f ). Assertion

(3.5.8) now follows.

3.5.4 Gibbs phenomenon

It is not reasonable to expect that the Fourier series of a discontinuous function

converges uniformly in a neighborhood of a discontinuity. The lack of uniformity in

the convergence can be measured in terms of the worst jump, called the overshoot.

The exact form of nonuniform convergence is illustrated in the following example:

Example 3.5.6. Consider the function

h(t) =

⎧⎪⎨⎪⎩

12− t when 0 < t ≤ 1

2

0 when t = 0

− 12− t when − 1

2< t < 0.

(3.5.11)

Clearly h(t) is a function of bounded variation and is continuous except at the point

t = 0 at which it has a jump discontinuity. Since h is an odd function, its Fourier

coefficients are

h(m) =∫ 1/2

−1/2h(t)e−2πimt dt =−2i

∫ 1/2

0( 1

2− t)sin(2πmt)dt =− i

2mπ

when m = 0 and h(0) = 0. The partial sums of the Fourier series of h are

(h∗DN)(t) =−i

2π ∑|m|≤Nm=0

e2πimt

m.

Notice thatd

dt(h∗DN)(t) = ∑

|m|≤Nm=0

e2πimt = DN(t)−1 .

Then, if we define d(s) = 1sin(πs) −

1πs

, we can write

(h∗DN)(t) =∫ t

0DN(s)−1ds

=−t +∫ t

0

sin((2N +1)πs)

sin(πs)ds

=−t +∫ t

0d(s)sin((2N +1)πs)ds+

∫ t

0

sin((2N +1)πs)

πsds .

Notice that d(s) is continuous at zero and d(0)= 0, while lims→0d(s)

s= π

6; thus d is a

differentiable function on [0, 12] and d′(0) = π

6. Moreover, lims→0 d′(s) = d′(0), thus

d is continuously differentiable on [0, 12]. Additionally both d and d′ are nonnegative

and increasing on [0, 12], thus d′ ≤ 4

π ; see the hint of Exercise 3.1.5. It follows that

∫ t

0d(s)sin((2N +1)πs)ds =−cos((2N +1)πt)

(2N +1)πd(t)+

∫ t

0d′(s)

cos((2N +1)πs)

(2N +1)πds

and the preceding expression is bounded in absolute value by (d( 1

2 )π + 1

2

d′( 12 )π ) 1

2N+1.

We deduce that

(h∗DN)(t) =−t +1

π

∫ t

0

sin((2N +1)πs)

sds+O

( 1

2N +1

),


where O(

12N+1

)is a function bounded by 1

π1

2N+1. Consequently,

(h∗DN)(t) =−t +1

π

∫ (2N+1)πt

0

sin(s)

sds+O

( 1

2N +1

).

Hence for t ∈ (0,− 12] we have limN→∞(h∗DN)(t) = −t + 1

ππ2= 1

2− t as expected.

Analogously for t ∈ [− 12,0) we have limN→∞(h ∗DN)(t) = − 1

2− t. Also for t = 0,

limN→∞(h∗DN)(0) = 0. Thus the Fourier series of h at zero converges to the “fair”

value of the average of h(0+) and h(0−) which happens to be h(0) = 0.

To quantitatively estimate the nonuniformity of the convergence of (h ∗DN)(t)we note that

(h∗DN)(t)− ( 12− t) =

1

π

∫ (2N+1)πt

0

sin(s)

sds− 1

2+O

( 1

2N +1

).

Thus for all N = 1,2, . . . and t ∈ (0, 12] we have

(h∗DN)(t)−h(t)≤ Si(π)

π− 1

2+

1

π

1

2N +1≤ .08949 · · ·+ π−1

2N +1.

Also, for any sequence tN → 0+ we have

limsupN→∞

[(h∗DN)(tN)−h(tN)

]≤ Si(π)

π− 1

2= .08949 . . . , (3.5.12)

while if for each N we consider the value tN = 1/(2N +1), we obtain that

limsupN→∞

[(h∗DN)(tN)−h(tN)

]=

Si(π)

π− 1

2= .08949 . . . . (3.5.13)

−0.4 −0.2 0.2 0.4

−0.6

−0.4

−0.2

0.2

0.4

0.6

Fig. 3.3 The partial sums (h∗D40)(t) showing the overshoot of approximately 9% of the jump of

h at zero.

The quantity .08949 . . . is called the overshoot of the partial sums of the Fourier

series of the function h in a neighborhood of zero. See Figure 3.3.

We now examine the preceding phenomenon in the setting of functions of

bounded variation. These functions can be written as a differences of two increasing

functions, so they have countable sets of discontinuities. Suppose we are given a

function f ∈ L1(T1) of bounded variation, and for the sake of simplicity, let us con-

sider the situation where it has exactly one discontinuity, say at the point t0 ∈ T1.

Consider the function h defined in (3.5.11) and define

f0(t) =

(f (t0+)− f (t0−)

)h(t− t0)+

f (t0+)+ f (t0−)2

when t = t0,

f (t0) when t = t0.(3.5.14)

Now the function f − f0 is of bounded variation and is also continuous and satisfies

( f − f0)(t0) = 0. In view of Corollary 3.5.2, the Fourier series of f − f0 converges

uniformly to f − f0 and so the lack of uniformity of the convergence of the partial

sums of f is due to the presence of f0.

We express these observations as a theorem.

Theorem 3.5.7. (a) Let h be defined in (3.5.11). Then the set of accumulation points

of sets of the form (h∗DN)(tN)N∈Z+ , where tN ∈ [0,1/2], is the interval

[0, Si(π)

π

]=

[0,0.58949 . . .

].

In particular if tN → 0 such that NtN → 12, then

limN→∞

(h∗DN)(tN) =Si(π)π = 0.58949 . . . .

(b) Let f be a function of bounded variation on the circle with a single discontinuity

at the point t0, such that f (t0+)− f (t0−) > 0. Then the set of accumulation points

of sets of the form ( f ∗DN)(tN)N∈Z+ , where tN ∈ [t0, t0+δ ], for some δ > 0, is the

interval [f (t0+)+ f (t0−)

2, f (t0+)+ f (t0−)

2+ Si(π)

π

(f (t0+)− f (t0−)

)].

In particular if tN → t0+ such that N(tN − t0)→ 12, then

limN→∞

( f ∗DN)(tN) =f (t0+)+ f (t0−)

2+ Si(π)

π

(f (t0+)− f (t0−)

).

Proof. (a) Since h ≥ 0 on (0, 12] and (h ∗DN)(t)→ 1

2− t for 0 < t ≤ 1

2we have

that all accumulation points of sequences (h∗DN)(tN) are nonnegative. We showed

in (3.5.12) that all accumulation points of sequences (h ∗DN)(tN)− h(tN) are at

mostSi(π)π − 1

2; but h(tN) ≤ 1

2, when tN ∈ [0, 1

2], hence all accumulation points of

sequences (h∗DN)(tN) are at mostSi(π)π and thus contained in

[0, Si(π)

π

]. Also, 0 is

attained as the accumulation point of (h∗DN)(0) and the numberSi(π)π is attained as

the accumulation point of the sequence (h∗DN)(1

2N+1) as shown in (3.5.13); notice


that the same assertion is valid for any other sequence tN → 0 such that NtN → 12.

Now, since the functions (h∗DN)(t) are continuous, given any c in[0, Si(π)

π

], there

is a t ′N between 0 and 12N+1

such that (h ∗DN)(t′N) = c for all N; this shows that

the set of accumulation points of sequences of the form ( f ∗DN)(tN)N∈Z+ is the

interval[0, Si(π)

π

].

(b) To examine the behavior of f ∗DN near the point of a single jump discon-

tinuity t0 of f , we reduce matters to the preceding situation as alluded earlier, by

introducing the function f0 defined in (3.5.14). Then for a sequence tN converging

to t0 from the right we have

( f ∗DN)(tN)

= ( f − f0)∗DN(tN)+(

f (t0+)− f (t0−))(h∗DN)(tN− t0)+

f (t0+)+ f (t0−)2

=(

f (t0+)− f (t0−))[(h∗DN)(tN− t0)−h(tN− t0)

]

+(

f (t0+)− f (t0−))h(tN− t0)+( f − f0)∗DN(tN)+

f (t0+)+ f (t0−)2

.

Applying limsupN→∞ and using (3.5.12), we obtain as N → ∞

limsupN→∞

( f ∗DN)(tN)≤(

f (t0+)− f (t0−))( Si(π)

π − 12+ 1

2

)+ f (t0+)+ f (t0−)

2,

where we used the following consequence of Theorem 3.5.4

limsupN→∞

( f − f0)∗DN(tN) = ( f − f0)(t0) = 0 .

This shows that all accumulation points of sequences of the form ( f ∗DN)(tN)N∈Z+

are at most(

f (t0+)− f (t0−)) Si(π)

π + 12

(f (t0+)+ f (t0−)

). Also, since all accumu-

lation points of sequences (h ∗DN)(tN) are nonnegative, when tN lies to the right

of t0, it follows from the identity f = ( f − f0)+ f0 and (3.5.14) that all accumula-

tion points of ( f ∗DN)(tN)N∈Z+ are at least 12

(f (t0+)+ f (t0−)

). As in part (a)

the intermediate value theorem implies that every point in the interval having the

aforementioned endpoints is also an accumulation point for the sequence at hand.

Exercises

3.5.1. Let Si(t) be the sine function as defined in (3.5.3).

(a) Prove that |π2−Si(t)| ≤ 2

t.

(b) Show that Si(Nt)→ π2

uniformly in t ∈ [δ ,∞) for any δ > 0.

3.5.2. Show that the sine integral function has the following expansion

Si(x) =∞

∑k=1

(−1)kx2k+1

(2k+1)(2k+1)!.

3.5.3. Let L11(T

1) be the space of all differentiable functions on T1 whose deriva-

tives are integrable. Obtain the inclusions L11(T

1)⊆ BV (T1)⊆ L∞(T1) as follows:

(a) If f ∈ L11(T

1), then Var( f )≤ ‖ f ′‖L1 .

(b) If f ∈ BV (T1), then ‖ f‖L∞ ≤ Var( f )+ | f (0)|.

3.5.4. (a) Let ak ≥ 0, sN = ∑Nk=−N ak, and σN = 1

N+1(σ0 + · · ·+σN). Suppose that

σN → L < ∞ as N → ∞. Prove that sN → L as N → ∞.

(b) Apply the preceding result to show that if a complex-valued function h on T1 is

continuous in a neighborhood of 0 and h(m) ≥ 0 for all m ∈ Z, then h(0) ≥ 0 and

∑m∈Z h(m) = h(0) < ∞; i.e., the partial sums of the Fourier series of h converge at

zero.

3.5.5. Let h ∈ L1(T1), t0 ∈ T1, and 0 < δ < 1/2.

(a) Show that (h∗DN)(t0)→ L as N → ∞ if and only if

limM→∞

∫ δ

0

(h(t0− t)+h(t0 + t)

2−L

)sin(Mt)

tdt = 0 .

(b) Conclude that if an integrable function h on T1 satisfies

∫ δ

0

|h(t0− t)+h(t0 + t)−2L|t

dt < ∞ ,

then (h∗DN)(t0)→ L as N → ∞.

(c) In particular, if there are constants C,β > 0 with β < 1 such that for all t with

0 < t < δ we have

|h(t0− t)+h(t0 + t)−2h(t0)| ≤Ctβ ,

then (h∗DN)(t0)→ h(t0) as N → ∞.

(d) If h is an odd function, then (h∗DN)(0)→ 0 as N → ∞.

3.5.6. Let f ∈ L1(T1) and suppose that (a,b) is an interval in T1. Then we have

limN→∞

∫ b

a( f ∗DN)(t)dt =

∫ b

af (t)dt .

[Hint: Use Theorems 3.5.4 and 3.5.5 and the fact that the operator f → f ∗DN is

self-adjoint.]

3.6 Lacunary Series and Sidon Sets

Lacunary series provide examples of 1-periodic functions on the line that possess

certain remarkable properties.

3.6 Lacunary Series 227

3.6.1 Definition and Basic Properties of Lacunary Series

We begin by defining lacunary sequences.

Definition 3.6.1. A sequence of positive integers Λ = λk∞k=1 is called lacunary if

there exists a constant A > 1 such that λk+1 ≥ Aλk for all k ∈ Z+.

Examples of lacunary sequences are provided by exponential sequences, such as

λk = 2k,3k,4k, . . . . Observe that polynomial sequences such as λk = 1+ k2 are not

lacunary. Note that lacunary sequences tend to infinity as k→ ∞.

An important observation about lacunary sequences is the following: for any

m,k0 ∈ Z+ we have

1≤ |m−λk0|< (1−A−1)λk0

=⇒ m /∈Λ . (3.6.1)

Indeed, to prove this assertion, notice that the closest numbers to λk0among of the

terms of the sequence λk∞k=1 are λk0+1 and λk0−1 (the latter only if k0 > 1) and

thus if j > k0 we have

|λ j−λk0| ≥ λk0+1−λk0

≥ Aλk0−λk0

= (A−1)λk0≥ (1−A−1)λk0

,

while if j < k0

|λ j−λk0| ≥ λk0

−λk0−1 ≥ λk0− 1

Aλk0

= (1−A−1)λk0.

Thus (3.6.1) follows.

We begin with the following result.

Proposition 3.6.2. Let λk∞k=1 be a lacunary sequence and let f be an integrable

function on the circle that is differentiable at a point and has Fourier coefficients

f (m) =

am when m = λk ,

0 when m = λk .(3.6.2)

Then we have

limk→+∞

f (λk)λk = 0 .

Proof. Applying translation, we may assume that the point at which f is differen-

tiable is the origin. Replacing f by the 1-periodic function

g(t) = f (t)− f (0)cos(2πt)− f ′(0)sin(2πt)

2π

we may assume that f (0) = f ′(0) = 0. (We have g(m) = f (m) for |m| ≥ 2 and thus

the final conclusion for f is equivalent to that for g.)

Using (3.6.1) and (3.6.2), we obtain that for any m ∈ Z we have

1≤ |m−λk|< (1−A−1)λk =⇒ f (m) = 0 . (3.6.3)

Let [t] denote the integer part of t. Given ε > 0, pick a positive integer k0 such that

if [(1−A−1)λk0] = 2N0, then N−2

0 < ε , and

sup

|x|<N− 1

40

∣∣∣∣f (x)

x

∣∣∣∣< ε . (3.6.4)

The expression in (3.6.4) can be made arbitrarily small, since f is differentiable at

the origin. Now take an integer k with k≥ k0 and set 2N = [min(A−1,1−A−1)λk],which is of course at least 2N0. Using (3.6.3), we obtain that for any trigonometric

polynomial KN of degree 2N with KN(0) = 1 we have

f (λk) =

∫

|x|≤ 12

f (x)KN(x)e−2πiλkx dx . (3.6.5)

We take KN = (FN/‖FN‖L2)2, where FN is the Fejer kernel. Using (3.1.18), we obtain

first the identity

∥∥FN

∥∥2

L2 =N

∑j=−N

(1− | j|

N +1

)2

= 1+1

3

N(2N +1)

N +1>

N

3(3.6.6)

and also the estimate

FN(x)2 ≤

(1

N +1

1

4x2

)2

, (3.6.7)

which is valid for |x| ≤ 1/2. In view of (3.6.6) and (3.6.7), we have the estimate

KN(x)≤3

16

1

N3

1

x4. (3.6.8)

We now use (3.6.5) to obtain

λk f (λk) = λk

∫

|x|≤ 12

f (x)KN(x)e−2πiλkx dx = I1

k + I2k + I3

k ,

where

I1k = λk

∫

|x|≤N−1f (x)KN(x)e

−2πiλkx dx ,

I2k = λk

∫

N−1<|x|≤N− 1

4

f (x)KN(x)e−2πiλkx dx ,

I3k = λk

∫

N− 1

4 <|x|≤ 12

f (x)KN(x)e−2πiλkx dx .

Since ‖KN‖L1 = 1, it follows that

|I1k | ≤

λk

Nsup

|x|<N−1

∣∣∣∣f (x)

x

∣∣∣∣≤(2N +1)ε

min(A−1,1−A−1)N,

which can be made arbitrarily small if ε is small. Also, using (3.6.8), we obtain

|I2k | ≤

3λk

16N3sup

|x|<N− 1

4

∣∣∣∣f (x)

x

∣∣∣∣∫

N−1<|x|≤N− 1

4

dx

|x|3 ≤3λk

16Nsup

|x|<N− 1

4

∣∣∣∣f (x)

x

∣∣∣∣ ,

which, as observed, is bounded by a constant multiple of ε . Finally, using again

(3.6.8), we obtain

|I3k | ≤

3

16N3Nλk

∫

N− 1

4 <|x|≤ 12

| f (x)|dx≤ 3

16N2

∥∥ f∥∥

L1 <3ε

16

∥∥ f∥∥

L1 .

It follows that for all k ≥ k0 we have

|λk f (λk)| ≤ |I1k |+ |I2

k |+ |I3k | ≤C( f )ε

for some fixed constant C( f ). This proves the required conclusion.

Corollary 3.6.3. (Weierstrass) There exists a continuous function on the circle that

is nowhere differentiable.

Proof. Consider the 1-periodic function

f (t) =∞

∑k=0

2−ke2πi3kt .

Since this series converges absolutely and uniformly, f is a continuous function. If f

were differentiable at a point, then by Proposition 3.6.2 we would have that 3k f (3k)

tends to zero as k→∞. Since f (3k) = 2−k for k≥ 0, this is not the case. Therefore, f

is nowhere differentiable. The real and imaginary parts of this function are displayed

in Figure 3.4.

3.6.2 Equivalence of Lp Norms of Lacunary Series

We now turn to one of the most important properties of lacunary series, equivalence

of their norms. It is a remarkable result that lacunary Fourier series have comparable

Lp norms for 1≤ p < ∞. More precisely, we have the following theorem:

Theorem 3.6.4. Let 1 ≤ λ1 < λ2 < λ3 < · · · be a lacunary sequence with constant

A > 1. SetΛ = λk : k ∈ Z+. Then for all 1≤ p <∞, there exists a constant Cp(A)

such that for all f ∈ L1(T1), with f (k) = 0, when k ∈ Z\Λ we have

-0.4 -0.2 0.2 0.4

-2

-1

1

2

-0.4 -0.2 0.2 0.4

-1

-0.5

0.5

1

Fig. 3.4 The graph of the real and imaginary parts of the function f (t) = ∑∞k=0 2−ke2πi3kt .

∥∥ f∥∥

Lp(T1)≤Cp(A)

∥∥ f∥∥

L1(T1). (3.6.9)

Moreover, the converse inequality to (3.6.9) is valid, and thus all Lp norms of lacu-

nary Fourier series are equivalent for 1≤ p < ∞.

Proof. We suppose initially that f ∈ L2(T1) and f is nonzero. We define

fN(x) =N

∑j=1

f (λ j)e2πiλ jx . (3.6.10)

Given 2 ≤ p < ∞, we pick an integer m with 2m > p and we also pick a positive

integer r such that Ar > m. Then we can write fN as a sum of r functions ϕs, s =1,2, . . . ,r, where each ϕs has Fourier coefficients that vanish except possibly on the

lacunary set

λkr+s : k ∈ Z+∪0= µ1,µ2,µ3, . . . .It is a simple fact that the sequence µkk is lacunary with constant Ar. Then we

have

∫ 1

0|ϕs(x)|2m dx = ∑

1≤ j1,..., jm,k1,...,km≤Nµ j1

+···+µ jm=µk1+···+µkm

ϕs(µ j1) · · · ϕs(µ jm)ϕs(µk1) · · · ϕs(µkm

) .

We claim that if µ j1 + · · ·+µ jm = µk1+ · · ·+µkm

, then

max(µ j1 , . . . ,µ jm) = max(µk1, . . . ,µkm

) .

Indeed, if max(µ j1 , . . . ,µ jm)> max(µk1, . . . ,µkm

), then

max(µ j1 , . . . ,µ jm)≤ µk1+ · · ·+µkm

≤ mmax(µk1, . . . ,µkm

) .

But since

Ar max(µk1, . . . ,µkm

)≤max(µ j1 , . . . ,µ jm) ,

it would follow that Ar ≤ m, which contradicts our choice of r. Likewise, we elimi-

nate the case max(µ j1 , . . . ,µ jm)< max(µk1, . . . ,µkm

). We conclude that these num-

bers are equal. We can now continue the same reasoning using induction to conclude

that if µ j1 + · · ·+µ jm = µk1+ · · ·+µkm

, then

µk1, . . . ,µkm

= µ j1 , . . . ,µ jm .

Using this fact in the evaluation of the previous multiple sum, we obtain

∫ 1

0|ϕs(x)|2m dx =

N

∑j1=1

· · ·N

∑jm=1

|ϕs(µ j1)|2 · · · |ϕs(µ jm)|2 =(∥∥ϕs

∥∥2

L2

)m,

which implies that ‖ϕs‖L2m = ‖ϕs‖L2 for all s ∈ 1,2, . . . ,r. Then we have

∥∥ fN

∥∥Lp ≤

∥∥ fN

∥∥L2m ≤

√r( r

∑s=1

∥∥ϕs

∥∥2

L2m

) 12=√

r( r

∑s=1

∥∥ϕs

∥∥2

L2

) 12=√

r∥∥ fN

∥∥L2 ,

since the functions ϕs are orthogonal in L2. Since r can be chosen to be [logA m]+1

and m can be taken to be [ p2]+1, we have now established the inequality

∥∥ fN

∥∥Lp(T1)

≤ cp(A)∥∥ fN

∥∥L2(T1)

, p≥ 2 , (3.6.11)

with cp(A) =√

1+[

logA

([ p

2]+1

)]for every fN of the form (3.6.10).

To replace fN by f in (3.6.11), we recall our assumption that f ∈ L2(T1). We

observe that fN → f in L2 and thus fN jtends to f a.e. for some subsequence. Then

Fatou’s lemma and (3.6.11) imply for 1 < p < ∞

∫ 1

0| f (x)|p dx =

∫ 1

0liminf

j→∞| fN j

(x)|p dx

≤ liminfj→∞

∫ 1

0| fN j

(x)|p dx

≤ cp(A)p liminf

j→∞

∥∥ fN j

∥∥p

L2

= cp(A)p∥∥ f

∥∥p

L2 .

We conclude that

∥∥ f∥∥

Lp(T1)≤ cp(A)

∥∥ f∥∥

L2(T1), p≥ 2 . (3.6.12)

By interpolation we obtain

∥∥ f∥∥

L2 ≤∥∥ f

∥∥ 23

L4

∥∥ f∥∥ 1

3

L1 ≤([logA 3]+1

) 12 · 23 ∥∥ f

∥∥ 23

L2

∥∥ f∥∥ 1

3

L1 .

We are assuming that 0 < ‖ f‖L2 < ∞ and the preceding inequality implies that

∥∥ f∥∥

L2(T1)≤

([logA 3]+1

)∥∥ f∥∥

L1(T1). (3.6.13)

Finally an easy consequence of Holder’s inequality is that

∥∥ f∥∥

Lp(T1)≤

∥∥ f∥∥

L2(T1), 1≤ p < 2 . (3.6.14)

Combining (3.6.12) and (3.6.14) with (3.6.13) yields (3.6.9) with

Cp(A) = cp(A)([logA 3]+1

)

for all 1 ≤ p < ∞ under the hypothesis that f (k) = 0 for all k ∈ Z \Λ and the

additional assumption that f ∈ L2.

We now extend the result to f ∈ L1(T1). Given f ∈ L1(T1) with f (k) = 0 when

k ∈ Z\Λ , consider the functions f ∗FM , where FM is the Fejer kernel and M ∈ Z+.

Then f ∗FM lie in L2, f ∗FM converge to f in L1 and in Lp, and f ∗FM(k) = 0 when

k ∈ Z\Λ . The inequality

∥∥ f ∗FM

∥∥Lp(T1)

≤Cp(A)∥∥ f ∗FM

∥∥L1(T1)

(3.6.15)

holds since f ∗FM lie in L2, so letting M → ∞ yields (3.6.9).

Theorem 3.6.4 describes the equivalence of the Lp norms of lacunary Fourier

series for p < ∞. The question that remains is whether there is a similar charac-

terization for the L∞ norms of lacunary Fourier series. Such a characterization is

investigated below. Before we state and prove this theorem, we need a classical tool,

referred to as a Riesz product.

Definition 3.6.5. A Riesz product is a function of the form

PN(x) =N

∏j=1

(1+a j cos(2πλ jx+2πγ j)

), (3.6.16)

where N is a positive integer, λ1 < λ2 < · · ·< λN is a lacunary sequence of positive

integers, a j are real numbers in [−1,1], and γ j ∈ [0,1].

We make a few observations about Riesz products. A simple calculation gives

that if PN, j(x) = 1+a j cos(2πλ jx+2πγ j), then

PN, j(m) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

1 when m = 0,12a je

2πiγ j when m = λ j,12a je

−2πiγ j when m =−λ j,

0 when m /∈ 0⋃∞j=1λ j,−λ j.

(3.6.17)

Assume that the constant A associated with the lacunary sequence λ1<λ2<· · ·<λN

satisfies A≥ 3. Then each integer m has at most one representation as a sum

m = ε1λ1 + · · ·+ εNλN ,

where ε j ∈ −1,1,0; see Exercise 3.6.1. We now calculate the Fourier coefficients

of the Riesz product defined in (3.6.16). For a fixed integer b, let us denote by δb the

sequence of integers that is equal to 1 at b and zero otherwise. Then, using (3.6.17),

we obtain that

PN, j = δ0 +12a je

2πiγ jδλ j+ 1

2a je

−2πiγ jδ−λ j,

and thus PN is the N-fold convolution of these functions. Using that δa ∗δb = δa+b,

we obtain

PN(m) =

⎧⎪⎨⎪⎩

1 when m = 0,

∏Nj=1

12a je

2πiε jγ j when m = ∑Nj=1 ε jλ j and ∑N

j=1 |ε j|> 0,

0 otherwise.

It follows that PN(λ j) = 0 since for j ≥ N + 1, λ j cannot be expressed as a linear

combination of λ1, . . . ,λN with coefficients in ±1,0. Also PN(λk) =12a je

2πiγk for

1≤ k ≤ N, since each λk is written uniquely as 0 ·λ1 + · · ·+0 ·λk−1 +1 ·λk. Hence

when A ≥ 3 we have that PN(λk) =12a je

2πiγk when 1 ≤ k ≤ N and PN(λk) = 0 for

k ≥ N +1.

Next, we discuss an important property of Riesz products. Suppose that for some

m ∈ Z we have PN(m) = 0. We write m = ∑Nj=1 ε jλ j uniquely with ε j ∈ −1,0,1.

Let k be the largest integer less than or equal to N such that εk = 0. Then we have

∣∣|m|−λk

∣∣≤ λ1 + · · ·+λk−1 ≤λk

Ak−1+ · · ·+ λk

A≤ λk

A

1

1− 1A

=λk

A−1. (3.6.18)

Another important property of the Riesz product is that since PN ≥ 0 we have

‖PN‖L1 =∫

T1PN(t)dt = PN(0) = 1 .

We recall the space A(T1) of all functions with absolutely summable Fourier

coefficients normed with the ℓ1 norm of the coefficients.

Theorem 3.6.6. Let 1 < λ1 < λ2 < λ3 < · · · be a lacunary sequence of integers with

constant A > 1. Set Λ = λk : k ∈ Z+. Then there exists a constant C(A) such that

for all f ∈ L∞(T1) with f (k) = 0 when k ∈ Z\Λ we have

∥∥ f∥∥

A(T1)= ∑

k∈Λ| f (k)| ≤C(A)

∥∥ f∥∥

L∞(T1). (3.6.19)

Proof. Let us assume first that A ≥ 3. Also fix f ∈ L∞(T1). We consider the Riesz

product

PN(x) =N

∏j=1

(1+ cos(2πλ jx+2πγ j)

),

where γ j is chosen to satisfy the identity | f (λ j)|= e2πiγ j f (λ j). In view of Parseval’s

relation and of the fact that ‖PN‖L1 = 1 we obtain

∣∣∣∣ ∑m∈Z

PN(m) f (m)

∣∣∣∣=∣∣∣∣∫ 1

0PN(x) f (x)dx

∣∣∣∣≤∥∥ f

∥∥L∞

, (3.6.20)

and the sum in (3.6.20) is finite, since the Fourier coefficients of PN form a finitely

supported sequence. But f (m) = 0 for m /∈Λ , while PN(λ j) =12e2πiγ j for 1≤ j≤ N

since A ≥ 3, and moreover, PN(λ j) = 0 for j ≥ N + 1, as observed earlier. Thus

(3.6.20) reduces to

1

2

N

∑j=1

∣∣ f (λ j)∣∣=

∣∣∣∣N

∑j=1

1

2e2πiγ j f (λ j)

∣∣∣∣≤∥∥ f

∥∥L∞

.

Letting N →∞, we deduce that ∑∞j=1 | f (λ j)| ≤ 2‖ f‖L∞ , which proves (3.6.19) when

A≥ 3.

We now consider the case A < 3. We fix 1 < A < 3 and we pick a positive integer

r such that

Ar > 3 and1

Ar−1< 1− 1

A. (3.6.21)

This is possible, since (Ar−1)−1 → 0 as r→ ∞.

For each s ∈ 1, . . . ,r, define the sequences λ sk = λs+(k−1)r indexed by k =

1,2,3, . . . and observe that λ s(k+1) > Arλ s

k for all k = 1,2, . . . ; i.e., each such se-

quence is lacunary with constant Ar. We consider the Riesz product

PsN(x) =

N

∏k=1

(1+ cos(2πλ s

k x+2πγsk)),

where γsk is defined via the identity | f (λ s

k )|= e2πiγsk f (λ s

k ).

Using (3.6.18) we obtain that, if m ∈ Z is such that PsN(m) = 0, then there exists

a k ∈ 1,2 . . . ,N such that

∣∣|m|−λ sk

∣∣< λ sk

Ar−1.

This combined with (3.6.21) yields

∣∣|m|−λ sk

∣∣<(

1− 1

A

)λ s

k .

Using (3.6.1) we obtain that either m =±λ sk or |m| /∈Λ . Thus we have

m ∈ Z+ : PsN(m) = 0 λ s

1 ,λs2 , . . . ,λ

sN∪Λ c .


This observation, the fact that f is supported in Λ , and Parseval’s relation yield

∣∣∣∣N

∑k=1

PsN(λ

sk ) f (λ s

k )

∣∣∣∣=∣∣∣∣ ∑

m∈Z

PsN(m) f (m)

∣∣∣∣=∣∣∣∣∫ 1

0Ps

N(x) f (x)dx

∣∣∣∣≤∥∥ f

∥∥L∞

. (3.6.22)

Since PN(λsk ) =

12e2πiγs

k for 1≤ k ≤ N, (3.6.22) reduces to

1

2

N

∑k=1

∣∣ f (λ sk )∣∣≤

∥∥ f∥∥

L∞.

Letting N → ∞ gives∞

∑j=1

∣∣ f (λ sj )∣∣≤ 2

∥∥ f∥∥

L∞.

Summing over s in the set 1,2, . . . ,r, we obtain the required conclusion with

C(A) = 2r and note that r can be taken to be [max(logA2A−1A−1

, log3 A)]+2.

Corollary 3.6.7. Let Λ = λk : k ∈ Z+ be a lacunary set and let f be a bounded

function on the circle that satisfies f (k) = 0 when k ∈ Z \Λ . Then f is almost ev-

erywhere equal to the absolutely (and uniformly) convergent series

f (x) = ∑k∈Λ

f (k)e2πikx a.e. (3.6.23)

and thus it is almost everywhere equal to a continuous function.

Proof. It follows from Theorem 3.6.6 that if f (k) = 0 when k ∈ Z\Λ , then we have

that f ∈ A(T1). Applying the inversion result in Proposition 3.2.5 we obtain that f is

almost everywhere equal to a continuous function and that (3.6.23) holds for almost

all x ∈ T1.

3.6.3 Sidon sets

Given a subset E of the integers, we denote by CE the space of all continuous func-

tions on T1 such that

m ∈ Z\E =⇒ f (m) = 0 . (3.6.24)

It is straightforward that CE is a closed subspace of all bounded functions on the

circle T1 with the standard L∞ norm.

Definition 3.6.8. A set of integers E is called a Sidon set if every function in CE has

an absolutely convergent Fourier series.

There are several characterizations of Sidon sets. We state them below.

Proposition 3.6.9. The following assertions are equivalent for a subset E of Z.


(1) There is a constant K such that for all trigonometric polynomials P with P

supported in E we have

∑m∈Z

|P(m)| ≤ K∥∥P

∥∥L∞

(2) There exists a constant K such that

∥∥ f∥∥ℓ1(Z)

≤ K∥∥ f

∥∥L∞(T1)

for every bounded function f on T1 with f supported in E.

(3) Every function f in CE has an absolutely convergent Fourier series; i.e., E is a

Sidon set.

(4) For every bounded function b on E there is a finite Borel measure µ on T1 such

that µ(m) = b(m) for all m ∈ E.

(5) For every function b on Z with the property b(m)→ 0 as m → ∞, there is a

function g ∈ L1(T1) such that g(m) = b(m) for all m ∈ E.

Proof. Suppose that (1) holds. Given f in L∞(T1) with f is supported in E, write

( f ∗FN)(x) =N

∑m=−N

(1− |m|

N +1

)f (m)e2πimx ,

where FN is the Fejer kernel. These are trigonometric polynomials whose Fourier

coefficients vanish on Z\E. Applying (1) we obtain

∑k∈Z

(1− |m|

N +1

)| f (m)| ≤ K

∥∥ f ∗FN

∥∥L∞

.

Letting N → ∞ we obtain (2).

It is trivial that (2) implies (3).

If (3) holds, then the map f → f is a linear bijection from CE to ℓ1(E). Moreover

its inverse mapping f → f is continuous, since

∥∥ f∥∥

L∞(T1)≤ sup

t∈[0,1]

∣∣∣∑k∈Z

f (k)e2πikt∣∣∣≤ ∑

k∈Z

| f (k)|=∥∥ f

∥∥ℓ1(Z)

.

By the open mapping theorem, it follows that f → f is a continuous mapping, which

proves the existence of a constant K such that (1) holds.

We have now proved the equivalence of (1), (2), and (3).

We show that (2) implies (4). If E is a Sidon set and if b is a bounded function

on E, say ‖b‖ℓ∞ ≤ 1, then the mapping

f → ∑m∈E

f (m)b(m)

is a bounded linear functional on CE with norm at most K. By the Hanh-Banach

theorem this functional admits an extension to C (T1) with the same norm. Hence

there is a measure µ , whose total variation ‖µ‖ does not exceed K, such that

∑m∈E

f (m)b(m) =∫

T1f (t)dµ(t) .

Taking f (t) = e2πimt in (2) we obtain µ(m) = b(m) for all m ∈ E.

If (4) holds and b(m) → 0 as |m| → ∞, using Lemma 3.3.2 there is a con-

vex sequence c(m) such that c(m) > 0, c(m) → 0 as |m| → ∞, c(−m) = c(m),and |b(m)| ≤ c(m) for all m ∈ Z. By (4), there is a finite Borel measure µ with

µ(m) = b(m)/c(m) for all m ∈ E.

By Theorem 3.3.4, there is a function g in L1(T1) such that g(m) = c(m) for all

m ∈ Z. Then b(m) = g(m)µ(m) for all m ∈ E. Since f = g ∗ µ is in L1, we have

b(m) = f (m) for all m ∈ E, and thus (4) implies (5).

Finally, if (5) holds, we show (3). Given f ∈ CE , we show that for an arbi-

trary sequence dm tending to zero, we have ∑m∈Z | f (m)dm| < ∞; this implies that

∑m∈Z | f (m)| < ∞. Given a sequence dm → 0, pick a function g in L1 such that

g(m) f (m) = | f (m)| |dm| for all m ∈ E by assumption (5). Then the series

∑m∈Z

g(m) f (m) = ∑m∈Z

f ∗g(m) (3.6.25)

has nonnegative terms and the function f ∗ g is continuous, thus FN ∗ ( f ∗ g)(0)→( f ∗g)(0) as N → ∞. It follows that DN ∗ ( f ∗g)(0)→ ( f ∗g)(0), thus the series in

(3.6.25) converges (see Exercise 3.5.4) and hence ∑m∈Z | f (m)dm|< ∞.

Example 3.6.10. Every lacunary set is a Sidon set. Indeed, suppose that E is a lacu-

nary set with constant A. If f is a continuous function which satisfies (3.6.24), then

Theorem 3.6.6 gives that

∑m∈Λ

| f (m)| ≤C(A)∥∥ f

∥∥L∞

< ∞ ;

hence f has an absolutely convergent Fourier series.

Example 3.6.11. There exist subsets of Z that are not Sidon. For example, Z \ 0is not a Sidon set. See Exercise 3.6.7.

Exercises

3.6.1. Suppose that 0 < λ1 < λ2 < · · · < λN is a lacunary sequence of integers

with constant A≥ 3. Prove that for every integer m there exists at most one N-tuple

(ε1, . . . ,εN) with each ε j ∈ −1,1,0 such that

m = ε1λ1 + · · ·+ εNλN .

[Hint: Suppose there exist two such N-tuples. Pick the largest k such that the coeffi-

cients of λk are different.]

3.6.2. Is the sequence λk =[e(logk)2]

, k = 2,3,4, . . . lacunary?

3.6.3. Let ak ≥ 0 for all k ∈ Z+ and 1 ≤ p < ∞. Show that there exist constants

Cp,cp such that for all N ∈ Z+ we have

cp

( N

∑k=1

|ak|2) 1

2 ≤(∫ 1

0

∣∣∣N

∑k=1

ake2πi2kx∣∣∣

p

dx

) 1p

≤Cp

( N

∑k=1

|ak|2) 1

2,

while

supx∈[0,1]

∣∣∣N

∑k=1

ake2πi2kx∣∣∣=

N

∑k=1

|ak| .

3.6.4. Suppose that 0< λ1 < λ2 < · · · is a lacunary sequence and let f be a bounded

function on the circle that satisfies f (m)= 0 whenever m∈Z\λ1,λ2, . . .. Suppose

also that

supt =0

| f (t)− f (0)||t|α = B < ∞

for some 0 < α < 1.

(a) Prove that there is a constant C such that | f (λk)| ≤CBλ−αk for all k ≥ 1.

(b) Prove that f ∈.Λα(T

1).[Hint: Let 2N = [(1−A−1)λk] and let KN be as in the proof of Proposition 3.6.2.

Write

f (λk) =

∫

|x|≤N−1( f (x)− f (0))e−2πiλkxKN(x)dx

+∫

N−1≤|x|≤ 12

( f (x)− f (0))e−2πiλkxKN(x)dx .

Use that ‖KN‖L1 = 1 and also the estimate (3.6.7). Part (b): Use the estimate in

part (a).]

3.6.5. Let f be an integrable function on the circle whose Fourier coefficients van-

ish outside a lacunary set Λ = λ1,λ2,λ3, . . .. Suppose that f vanishes identically

in a small neighborhood of the origin. Show that f is in C ∞(T1).[Hint: Let 2N = [(1−A−1)λk] and let KN be as in the proof of Proposition 3.6.2.

Write

f (λk) =∫

|x|≤ 12

f (x)e−2πiλkxKN(x)dx

and use estimate (3.6.7) to obtain that f is in C 2. Continue by induction.]

3.6.6. Let 1 < a,b < ∞. Consider the 1-periodic function

f (x) =∞

∑k=0

a−ke2πibkx .

Prove that the following statements are equivalent:

(a) f is differentiable at a point.

(b) b < a.

(c) f is differentiable everywhere.

3.6.7. Use the example in Proposition 3.4.6 (a) to show that Z\q1, . . . ,qL is not

a Sidon set for any finite subset q1, . . . ,qL of the integers.

3.6.8. Let 0 < δ < 1. Let E be a subset of the integers such that for any sequence

of complex numbers dmm∈E with |dm|= 1 there is a finite Borel measure µ on T1

such that

|µ(m)−dm|< 1−δfor all m ∈ E. Show that E is a Sidon set.[Hint: Given f be in CE define dm via the identity dm f (m) = | f (m)| if f (m) =

0, otherwise set dm = 1. For the measure µ given by the hypothesis, notice that

Re(µ(m) f (m))≥ δ | f (m)| for all m ∈ Z.]

HISTORICAL NOTES

Trigonometric series in one dimension were first considered in the study of the vibrating string

problem and are implicitly contained in the work of d’Alembert, D. Bernoulli, Clairaut, and Euler.

The analogous problem for vibrating higher-dimensional bodies naturally suggested the use of mul-

tiple trigonometric series. However, it was the work of Fourier on steady-state heat conduction that

inspired the subsequent systematic development of such series. Fourier announced his results in

1811, although his classical book Theorie de la chaleur was published in 1822. This book contains

several examples of heuristic use of trigonometric expansions and motivated other mathematicians

to carefully study such expansions. The systematic development of the theory of Fourier series

began by Dirichlet [96], who studied the pointwise convergence of the Fourier series of piecewise

monotonic functions via the use of the kernel DN , today called the Dirichlet kernel.

The fact that the Fourier series of a continuous function can diverge was first observed by

DuBois Reymond in 1873. The Riemann–Lebesgue lemma was first proved by Riemann in his

memoir on trigonometric series (appeared between 1850 and 1860). It carries Lebesgue’s name

today because Lebesgue later extended it to his notion of integral. The rebuilding of the theory

of Fourier series based on Lebesgue’s integral was mainly achieved by de la Vallee-Poussin and

Fatou.

Theorem 3.3.16 was obtained by Bernstein [26] in dimension n = 1. Higher-dimensional ana-

logues of the Hardy–Littlewood series of Exercise 3.3.8 were studied by Wainger [370]. These

series can be used to produce examples indicating that the restriction s > α + n/2 in Bernstein’s

theorem is sharp even in higher dimensions. Part (b) of Theorem 3.4.4 is due to Lebesgue when

n = 1 and Marcinkiewicz and Zygmund [243] when n = 2. Marcinkiewicz and Zygmund’s proof

also extends to higher dimensions. The proof given here is based on Lemma 3.4.5 proved by Stein

[342] in a different context. The proof of Lemma 3.4.5 presented here was suggested by T. Tao.

Abel proved that if an infinite series ∑∞k=0 ak converges and has sum L, then the power series

f (x) = ∑∞k=0 akxk converges for |x| < 1 and tends to L as x → 1−. The converse of this theorem

under the additional assumption that kak → 0 as k→ ∞ was proved by Tauber [358]. Hardy [143]

extended Tauber’s result (Theorem 3.5.1) for Cesaro summability under the weaker assumption that

the sequence kak is bounded. Jordan [180] studied of functions of bounded variation and proved

Theorem 3.5.4. The existence of a continuous function which is nowhere differentiable (Corollary

3.6.3) was first published in 1872 by K. Weierstrass, although earlier findings of such functions

were published later. The exposition on Sidon sets is taken from the classical article of Rudin

[305], which also contains Exercise 3.6.8.


The Gibbs phenomenon (a version of Theorem 3.5.7) was discovered by Wilbraham [376] and

rediscovered by Gibbs [128]; this phenomenon describes the particular way in which the Fourier

sums of a piecewise continuously differentiable periodic function have large oscillations and over-

shoot at the jump discontinuity of the function. Bocher [29] gave a detailed mathematical analysis

of that overshoot, which he called the “Gibbs phenomenon”.

The main references for trigonometric series are the books of Bary [20] and Zygmund [388],

[389]. Other references for one-dimensional Fourier series include the books of Edwards [106],

Dym and McKean [105], Katznelson [190], Korner [202], Pinsky [283], and the first eight chapters

in Torchinsky [363]. The reader may also consult the book of Krantz [203] for a historical intro-

duction to the subject of Fourier series. A review of the heritage and continuing significance of

Fourier Analysis is written by Kahane [182].

A classical treatment of multiple Fourier series can be found in the last chapter of Bochner’s

book [32] and in parts of his other book [31]. Other references include the last chapter in Zygmund

[389], the books of Yanushauskas [381] (in Russian) and Zhizhiashvili [384], the last chapter in

Stein and Weiss [348], and the article of Alimov, Ashurov, and Pulatov in [3]. A brief survey article

on the subject was written by Ash [11]. More extensive expositions were written by Shapiro [320],

Igari [171], and Zhizhiashvili [383]. A short note on the history of Fourier series was written by

Zygmund [390]. The book of Shapiro [321] contains a very detailed study of Fourier series in

several variables as well as applications of this theory.

Chapter 4

Topics on Fourier Series

In this chapter we go deeper into the theory of Fourier series and we study topics

such as convergence in norm and the conjugate function, divergence of Fourier se-

ries and Bochner–Riesz summability. We also study transference of multipliers on

the torus and of maximal multipliers. This is a powerful technique that allows one

to infer results concerning Fourier series from corresponding results about Fourier

integrals and vice versa.

We also take a quick look at applications of Fourier series such as the isoperti-

metric inequality problem, the distribution of lattice points in a ball, and the heat

equation. The power of Fourier series techniques manifests itself in the study of

these problems which represent only a small part of the wide and vast range of ap-

plications of the subject known today.

4.1 Convergence in Norm, Conjugate Function,

and Bochner–Riesz Means

In this section we address the following fundamental question: Do Fourier series

converge in norm? We begin with some abstract necessary and sufficient conditions

that guarantee such a convergence. In one dimension, we are able to reduce matters

to the study of the so-called conjugate function on the circle, a sister operator of

the Hilbert transform, which is the center of study of the next chapter. In higher

dimensions the situation is more complicated, but we are able to give a positive

answer in the case of square summability.



241

242 4 Topics on Fourier Series

4.1.1 Equivalent Formulations of Convergence in Norm

The question we pose is for which indices p, with 1≤ p < ∞, we have

∥∥DnN ∗ f − f

∥∥Lp(Tn)

→ 0 as N → ∞ , (4.1.1)

and similarly for the circular Dirichlet kernel

DnN . We tackle this question by looking

at an equivalent formulation of it.

Theorem 4.1.1. For R > 0 and m ∈ Zn, let a(m,R) be complex numbers such that

(i) For every R > 0 there is a qR such that a(m,R) = 0 if |m|> qR.

(ii) There is an M0 < ∞ such that |a(m,R)| ≤M0 for all m ∈ Zn and all R > 0.

(iii) For each m∈Zn, the limit of a(m,R) exists as R→∞ and limR→∞ a(m,R) = am.

Let 1≤ p < ∞. For f ∈ Lp(Tn) and x ∈ Tn define

SR( f )(x) = ∑m∈Zn

a(m,R) f (m)e2πim·x

noting that the sum is well defined because of (i). Also, for h ∈ C ∞(Tn) define

A(h)(x) = ∑m∈Zn

amh(m)e2πim·x.

Then for all f ∈ Lp(Tn) the sequence SR( f ) converges in Lp as R→∞ if and only if

there exists a constant K < ∞ such that

supR>0

∥∥SR

∥∥Lp→Lp ≤ K. (4.1.2)

Furthermore, if (4.1.2) holds, then for the same constant K we have

suph∈C∞

h =0

∥∥A(h)∥∥

Lp∥∥h∥∥

Lp

≤ K , (4.1.3)

and then A extends to a bounded operator A from Lp(Tn) to itself; moreover, for

every f ∈ Lp(Tn) we have that SR( f )→ A( f ) in Lp as R→ ∞.

Proof. If SR( f ) converges in Lp, then ‖SR( f )‖Lp ≤ C f for some constant C f that

depends on f ∈ Lp(Tn). Moreover, each SR is a bounded operator from Lp(Tn) to

itself with norm at most #m ∈ Zn : |m| ≤ qRM0. Thus SRR>0 is a family of Lp-

bounded linear operators that satisfy supR>0 ‖SR( f )‖Lp ≤ C f for each f ∈ Lp(Tn).The uniform boundedness theorem applies and yields that the operator norms of SR

from Lp to Lp are bounded uniformly in R. This proves (4.1.2).

Conversely, assume (4.1.2). For h ∈ C ∞(Tn), we have that

limR→∞

∑m∈Zn

a(m,R)h(m)e2πim·x = ∑m∈Zn

amh(m)e2πim·x

4.1 Convergence in Norm, Conjugate Function, and Bochner–Riesz Means 243

in view of property (ii) and of the Lebesgue dominated convergence theorem, since

∑m∈Zn |h(m)|< ∞. Fatou’s lemma now gives

∥∥A(h)∥∥

Lp =∥∥ lim

R→∞SR(h)

∥∥Lp ≤ liminf

R→∞

∥∥SR(h)∥∥

Lp ≤ K∥∥h

∥∥Lp ;

hence (4.1.3) holds. Thus A extends to a bounded operator A on Lp(Tn) by density.

We show that for all f ∈ Lp(Tn) we have SR( f )→ A( f ) in Lp as R → ∞. Fix

f in Lp(Tn) and let ε > 0 be given. Pick a trigonometric polynomial P satisfying

‖ f −P‖Lp ≤ ε . Let d be the degree of P. Then there is an R0 > 0 such that for all

R > R0 we have

∑|m1|+···+|mn|≤d

|a(m,R)−am| |P(m)| ≤ ε

since a(m,R)→ am for every m with |m1|+ · · ·+ |mn| ≤ d. We deduce that

∥∥SR(P)−A(P)∥∥

Lp ≤∥∥SR(P)−A(P)

∥∥L∞

≤ ∑|m1|+···+|mn|≤d

|a(m,R)−am| |P(m)|

≤ ε ,

whenever R > R0. Then

∥∥SR( f )−A( f )∥∥

Lp ≤∥∥SR( f )−SR(P)

∥∥Lp +

∥∥SR(P)−A(P)∥∥

Lp +∥∥A(P)−A( f )

∥∥Lp

≤ Kε+ ε+Kε = (2K +1)ε

for R > R0. This proves that SR( f ) converges to A( f ) in Lp as R→ ∞.

The most interesting situation arises, of course, when a(m,R)→ am = 1 for all

m ∈ Zn. In this case A (and A) is the identity operator, and thus we expect the oper-

ators SR( f ) to converge back to f as R→ ∞. We should keep in mind the following

three examples:

(a) The sequence a(m,R) = 1 when |m j| ≤ R for all j ∈ 1,2, . . . ,n and zero other-

wise, in which case the operator SR of Theorem 4.1.1 is

SR( f ) = f ∗DnR ; (4.1.4)

(b) The sequence a(m,R) = 1 when |m| ≤ R and zero otherwise, in which case the

SR of Theorem 4.1.1 isSR( f ) = f ∗

Dn

R ; (4.1.5)

(c) The sequence a(m,R) =(1− |m|2

R2

)α+

, for some α > 0, in which case we denote

SR by BαR .

Definition 4.1.2. The Bochner–Riesz operator or Bochner–Riesz means of order

α ≥ 0 is the operator

BαR ( f )(x) = ∑m∈Zn

|m|≤R

(1− |m|2

R2

)αf (m)e2πim·x (4.1.6)

defined on integrable functions f on Tn.

Corollary 4.1.3. Let 1≤ p<∞ and α ≥ 0. Let SR andSR be as in (4.1.4) and (4.1.5),

respectively, and let BαR be the Bochner–Riesz means as defined in (4.1.6). Then

∀ f ∈ Lp(Tn), limR→∞

∥∥DnR ∗ f − f

∥∥Lp = 0 ⇐⇒ sup

R≥0

∥∥SR

∥∥Lp→Lp < ∞,


∥∥ DnR ∗ f − f

∥∥Lp = 0 ⇐⇒ sup

R≥0

∥∥ SR

∥∥Lp→Lp < ∞,


∥∥BαR ∗ f − f∥∥

Lp = 0 ⇐⇒ supR≥0

∥∥BαR∥∥

Lp→Lp < ∞.

Example 4.1.4. We investigate the one-dimensional case in some detail. We take

n = 1, and we define a(m,N) = 1 for all −N ≤ m ≤ N, and zero otherwise. Then

SN( f ) =SN( f ) = DN ∗ f , where DN is the Dirichlet kernel. Clearly, the expressions

‖SN‖Lp→Lp are bounded above by the L1 norm of DN , but this estimation yields a

bound that blows up as N → ∞. We later show, via a more delicate argument, that

the expressions ‖SN‖Lp→Lp are uniformly bounded in N when 1 < p < ∞.

This reasoning, however, allows us to deduce that for some function g ∈ L1(T1),SN(g) may not converge in L1. This is also a consequence of the proof of Theorem

4.2.1; see (4.2.13). Note that since the Fejer kernel FM has L1 norm 1, we have

∥∥SN

∥∥L1→L1 ≥ lim

M→∞

∥∥DN ∗FM

∥∥L1 =

∥∥DN

∥∥L1 .

This implies that the expressions ‖SN‖L1→L1 are not uniformly bounded in N, and

therefore Corollary 4.1.3 gives that for some f0 ∈ L1(T1), SN( f0) does not converge

to f0 in L1.

Although the partial sums of Fourier series fail to convergence in L1(Tn), it is a

consequence of Plancherel’s theorem that they converge in L2(Tn). More precisely,

if f ∈ L2(Tn), then

∥∥ Dn

N ∗ f − f∥∥2

L2 = ∑|m|>N

| f (m)|2 → 0

as N → ∞ and the same result is true for DnN ∗ f and for BαR ∗ f ; for the latter, we

apply Theorem 4.1.1, noting that

∥∥BαR ( f )∥∥2

L2 = ∑m∈Zn

∣∣∣∣(

1− |m|2

R2

)α

+

∣∣∣∣2

| f (m)|2 ≤ ∑m∈Zn

| f (m)|2 =∥∥ f

∥∥2

L2

and thus supR>0 ‖BαR‖L2→L2 ≤ 1.

Motivated by the preceding discussion for p = 2, it is natural to pose the follow-

ing question. Can 2 be replaced by p = 2 in the preceding results? This question has

an affirmative answer in dimension one for DN . In higher dimensions an interesting

dichotomy appears. As a consequence of the one-dimensional result, the square par-

tial sums DnN ∗ f converge in Lp to a given f in Lp(Tn), but for the circular partial

sums this may not be the case.

We begin the discussion with the one-dimensional situation.

Definition 4.1.5. For f ∈ C ∞(T1) define the conjugate function f by

f (x) =−i ∑m∈Z1

sgn(m) f (m)e2πimx,

where sgn(m) = 1 for m > 0, −1 for m < 0, and 0 for m = 0. Also define the Riesz

projections P+ and P− by

P+( f )(x) =∞

∑m=1

f (m)e2πimx , (4.1.7)

P−( f )(x) =−1

∑m=−∞

f (m)e2πimx . (4.1.8)

Observe that f = P+( f )+P−( f )+ f (0), while f = −iP+( f )+ iP−( f ), when f

is in C ∞(T1). Consequently, one has

P+( f ) =1

2( f + i f )− 1

2f (0) (4.1.9)

and therefore the Lp boundedness of the operator f → f is equivalent to that of the

operator f → P+( f ), since the identity and the operator f → f (0) are obviously

Lp bounded. Clearly, these statements are also valid for the other Riesz projection

f → P−( f ). The following is a consequence of Theorem 4.1.1.

Proposition 4.1.6. Let 1≤ p <∞. Then the expressions SN( f ) = DN ∗ f converge to

f in Lp(T1) as N → ∞ if and only if there exists a constant Cp > 0 such that for all

smooth functions f on Tn we have ‖ f ‖Lp(T1) ≤Cp‖ f‖Lp(T1).

Proof. In view of Corollary 4.1.3, the fact that for all f ∈ Lp(T1), SN( f )→ f in Lp

as N → ∞ is equivalent to the uniform (in N) Lp boundedness of SN .

We note the validity of the identity

e−2πiNx2N

∑m=0

(f (·)e2πiN(·)) (m)e2πimx =

N

∑m=−N

f (m)e2πimx .

Since multiplication by exponentials does not affect Lp norms, this identity implies

that the norm of the operator SN( f ) = DN ∗ f from Lp to Lp is equal to that of the

operator

S′N(g)(x) =2N

∑m=0

g(m)e2πimx

from Lp to Lp. Therefore,

supN≥0

∥∥SN

∥∥Lp→Lp < ∞ ⇐⇒ sup

N≥0

∥∥S′N∥∥

Lp→Lp < ∞ , (4.1.10)

and both of these statements are equivalent to the fact that for all f ∈ Lp(T1),SN( f )→ f in Lp as N → ∞.

We have already observed that the Lp boundedness of the conjugate function is

equivalent to that of P+. Therefore, it suffices to show that the Lp boundedness of

P+ is equivalent to the uniform Lp boundedness of S′N .

Suppose first that supN≥0 ‖S′N‖Lp→Lp <∞. Theorem 4.1.1 applied to the sequence

a(m,R) = 1 for 0≤m≤ R and a(m,R) = 0 otherwise gives that the operator A( f ) =

P+( f )+ f (0) is bounded on Lp(T1). Hence so is P+.

Conversely, suppose that P+ extends to a bounded operator from Lp(T1) to itself.

For all h in C ∞(Tn) we can write

S′N(h)(x) =∞

∑m=0

h(m)e2πimx−∞

∑m=2N+1

h(m)e2πimx

=∞

∑m=1

h(m)e2πimx + h(0)− e2πi(2N)x∞

∑m=1

h(m+2N)e2πimx

= P+(h)(x)− e2πi(2N)xP+(e−2πi(2N)(·)h)+ h(0) .

This identity implies that

supN≥0

∥∥S′N( f )∥∥

Lp ≤(2∥∥P+

∥∥Lp→Lp +1

)∥∥ f∥∥

Lp (4.1.11)

for all f smooth, and by density for all f ∈ Lp(T1). Note that S′N is well defined on

Lp(T1). Thus the operators S′N are uniformly bounded on Lp(Tn).Thus the uniform Lp boundedness of SN is equivalent to the uniform Lp bound-

edness of S′N , which is equivalent to the Lp boundedness of P+, which in turn is

equivalent to the Lp boundedness of the conjugate function.

4.1.2 The Lp Boundedness of the Conjugate Function

We know now that convergence of Fourier series in Lp is equivalent to the Lp bound-

edness of the conjugate function or either of the two Riesz projections. It is natural

to ask whether these operators are Lp bounded.

Theorem 4.1.7. Given 1 0 such that for all f in

C ∞(T1) we have ∥∥ f∥∥

Lp ≤ Ap

∥∥ f∥∥

Lp . (4.1.12)

Thus the operator f → f has a bounded extension on Lp(T1) that also satisfies

(4.1.12).

Consequently, the Fourier series of Lp functions on the circle converge back to

the functions in the Lp norm for 1 < p < ∞.

Proof. In proving the inequality (4.1.12), we make the following reductions:

(a) We assume that f is trigonometric polynomial.

(b) We assume that f (0) = 0.

(c) We assume that f is real valued.

Since f is a real-valued function, we have that f (−m) = f (m) for all m, and since

f (0) = 0, we can write

f (t) =−i∞

∑m=1

f (m)e2πimt + i∞

∑m=1

f (−m)e−2πimt = 2Re

[− i

∞

∑m=1

f (m)e2πimt

],

which implies that f is also real-valued (see also Exercise 4.1.4(b)). Therefore the

polynomial f + i f contains only positive frequencies. Thus for k ∈ Z+ we have

∫

T1( f (t)+ i f (t))2k dt = 0 .

Expanding the 2k power and taking real parts, we obtain

k

∑j=0

(−1)k− j

(2k

2 j

)∫

T1f (t)2k−2 j f (t)2 j dt = 0 ,

where we used that f is real-valued. Therefore,

∥∥ f∥∥2k

L2k ≤k

∑j=1

(2k

2 j

)∫

T1f (t)2k−2 j f (t)2 j dt

≤k

∑j=1

(2k

2 j

)∥∥ f∥∥2k−2 j

L2k

∥∥ f∥∥2 j

L2k ,

by applying Holder’s inequality with exponents 2k/(2k−2 j) and 2k/(2 j) to the jth

term of the sum. Dividing the last inequality by ‖ f‖2kL2k , we obtain

R2k ≤k

∑j=1

(2k

2 j

)R2k−2 j, (4.1.13)

where R = ‖ f ‖L2k/‖ f‖L2k . If R > 0 satisfies (4.1.13), then R≤C2k, where C2k is the

largest real root of the polynomial g(t) = t2k−∑kj=1

(2k2 j

)t2k−2 j. (Since g(0)< 0 and

limt→∞ g(t) =∞, g has at least one real root.) We conclude that if f satisfies (a), (b),

and (c), then we have for all k = 1,2, . . .

∥∥ f∥∥

L2k ≤C2k

∥∥ f∥∥

L2k . (4.1.14)

We now remove assumptions (a), (b), and (c). We first remove assumption (c).

Given a complex-valued trigonometric polynomial f with f (0) = 0, we write

f (t) =N

∑j=−N

c je2πi jt =

[N

∑j=−N

c j + c− j

2e2πi jt

]+ i

[N

∑j=−N

c j− c− j

2ie2πi jt

]

(with c0 = 0) and we note that the expressions inside the square brackets are real-

valued trigonometric polynomials. Thus we can express f as P+ iQ, where P and

Q are real-valued trigonometric polynomials, and applying (4.1.14) to P and Q we

obtain the inequality∥∥ f

∥∥L2k ≤ 2C2k

∥∥ f∥∥

L2k (4.1.15)

for all trigonometric polynomials f with f (0) = 0.

Next, we remove the assumption that f (0) = 0. We write f = ( f − f (0))+ f (0),we observe that the conjugate function of a constant is zero, and we apply (4.1.15)

to obtain

∥∥ f∥∥

L2k ≤ 2C2k

∥∥ f − f (0)∥∥

L2k ≤ 2C2k

[∥∥ f∥∥

L2k +‖ f‖L1

]≤ 4C2k

∥∥ f∥∥

L2k .

Since trigonometric polynomials are dense in Lp, it follows that the operator f → f

has a bounded extension on L2k that satisfies (4.1.12) for all f ∈ L2k, and in particular

for all f ∈ C ∞(T1).Every real number p ≥ 2 lies in an interval of the form [2k,2k + 2], for some

k ∈ Z+. Theorem 1.3.4 gives that for all 2≤ p < ∞ there is a constant Ap such that

∥∥ f∥∥

Lp ≤ Ap

∥∥ f∥∥

Lp (4.1.16)

when f is a simple function. Thus the conjugate function has a bounded extension

on Lp that satisfies (4.1.16) when p≥ 2.

To extend this result for p< 2 we use duality. We observe that the adjoint operator

of f → f is f → − f . Indeed, for f ,g in C ∞(T1) we have

〈 f |g〉= ∑m∈Z

−isgn(m) f (m)g(m) =− ∑m∈Z

f (m)−isgn(m)g(m) =−〈 f | g〉 .

By duality, estimate (4.1.16) is also valid for 1 < p≤ 2 with constant Ap′ = Ap.

We extend the preceding result to higher dimensions.

Theorem 4.1.8. Let 1 < p < ∞ and f ∈ Lp(Tn). Then DnN ∗ f converges to f in Lp

as N → ∞.

Proof. As a consequence of Corollary 4.1.3, Proposition 4.1.6, and Theorem 4.1.7,

it suffices to show that for all f trigonometric polynomials on Tn we have

supN>0

∫ 1

0· · ·

∫ 1

0

∣∣(DnN ∗ f )(x)

∣∣pdx1 · · ·dxn ≤ Knp

∥∥ f∥∥p

Lp(Tn).

Obviously, this inequality is valid in dimension n = 1. We extend it by induction

to all dimensions. We assume that it is valid in dimension n− 1 and we prove it in

dimension n.

Let x′= (x2, . . . ,xn)∈Tn−1. For a fixed trigonometric polynomial f , and for fixed

N ≥ 0 and x′ ∈ Tn−1, define a trigonometric polynomial gN,x′ on T1 by setting

gN,x′(x1) = ∑m1∈Z

[∑

|m2|,...,|mn|≤N

e2πim′·x′ f (m1,m′)

]e2πim1x1

where m′ = (m2, . . . ,mn). Then we have

gN,x′(x1) = ∑|m2|,...,|mn|≤N

e2πim′·x′[∑

m1∈Z

e2πim1x1 f (m1,m′)

]

= ∑|m2|,...,|mn|≤N

e2πim′·x′[∫

Tn−1f (x1,y

′)e−2πim′·y′dy′]

= ∑|m2|,...,|mn|≤N

e2πim′·x′ fx1(m′)

= (Dn−1N ∗ fx1

)(x′),

where fx1is the trigonometric polynomial of n− 1 variables defined by fx1

(x′) =f (x1,x

′). We also have that

(DN ∗gN,x′)(x1) = (DnN ∗ f )(x1,x

′) .

Combining this information, we write

∫

Tn−1

∫ 1

0

∣∣(DnN ∗ f )(x1,x

′)∣∣p

dx1 dx′

=

∫

Tn−1

∫ 1

0

∣∣(DN ∗gN,x′)(x1)∣∣p

dx1 dx′

≤ K p

∫

Tn−1

∫ 1

0

∣∣gN,x′(x1)∣∣p

dx1 dx′

= K p

∫ 1

0

∫

Tn−1

∣∣(Dn−1N ∗ fx1

)(x′)∣∣p

dx′ dx1


≤ K pK(n−1)p

∫

Tn−1

∫ 1

0

∣∣ fx1(x′)

∣∣pdx′ dx1

= Knp∥∥ f

∥∥p

Lp(Tn),

where the penultimate inequality follows from the induction hypothesis.

4.1.3 Bochner–Riesz Summability

In dimension 1 the Fejer means of an integrable function are better behaved than the

Dirichlet means. We investigate whether there is a similar phenomenon in higher

dimensions. Recall that the circular (or spherical) partial sums of the Fourier series

of f are given by

( f ∗Dn

R)(x) = ∑m∈Zn

|m|≤R

f (m)e2πim·x ,

where R≥ 0. Taking the averages of these expressions, we obtain

1

R

∫ R

0( f ∗

Dn

r )(x)dr = ∑m∈Zn

|m|≤R

(1− |m|

R

)f (m)e2πim·x = B1

R( f )(x) ,

and we call these expressions the circular Cesaro means (or circular Fejer means)

of f . It turns out that the circular Cesaro means of integrable functions on T2 always

converge in L1, but in dimension 3, this may fail. Theorem 4.2.5 gives an example of

an integrable function f on T3 whose circular Cesaro means diverge a.e. However,

we show below that this is not the case if the circular Cesaro means of a function f

in L1(T3) are replaced by the only slightly different-looking means

∑m∈Zn

|m|≤R

(1− |m|

R

)1+εf (m)e2πim·x ,

for some ε > 0. This discussion suggests that the preceding expressions behave bet-

ter as ε increases, but for a fixed ε they get worse as the dimension increases. The

need to understand the behavior of these operators for different values of α ≥ 0 led

to introduction of the operators BαR given in Definition 4.1.2.

The family of operators BαR forms a natural “spherical” analogue of the Cesaro–

Fejer sums. It turns out that there is no significant difference in the behavior of

these means if the expression(1− |m|2

R2

)αin (4.1.6) is replaced by the expression(

1− |m|R

)α; see Exercise 4.3.1. The advantage of the quadratic expression in (4.1.6)

is that it has an easily computable kernel and yields the elegant reproducing formula

BαR ( f ) =2Γ (α+1)

Γ (α−β )Γ (β +1)

1

R

∫ R

0

(1− r2

R2

)α−β−1(r2

R2

)β+ 12

Bβr ( f )dr , (4.1.17)

which precisely quantifies the way in which BαR is smoother than BβR when α > β .

Identity (4.1.17) also says that when α > β , the operator BαR ( f ) is an average of the

operators Bβr ( f ), 0 < r < R, with respect to a certain density.

Note that the Bochner–Riesz means of order zero coincide with the circular (or

spherical) Dirichlet means, and, as we have seen, these converge in L2(Tn). We

address an analogous question on Lp(Tn) for p = 2.

Proposition 4.1.9. Let 1 ≤ p < ∞ and f ∈ Lp(Tn). Then the Bochner–Riesz means

BαR ( f ) converge to f in Lp(Tn) as R→ ∞ when α > (n−1)∣∣ 1

p− 1

2

∣∣. Moreover, if f

is continuous on Tn and α > n−12

, then BαR ( f ) converges to f uniformly as R→ ∞.

Proof. For z ∈ C with Rez≥ 0, consider the function

mz(ξ ) = (1−|ξ |2)z+

defined for ξ in Rn. Note that ‖mz‖L∞ = 1. Using an identity proved in Appendix

B.5, we have that

(mz)∨(y) = Kz(y) =

Γ (z+1)

πz

J n2+z(2π|y|)|y| n

2+z, (4.1.18)

where y ∈Rn and Jν is the Bessel function of order ν . The estimates in Appendices

B.6 and B.7 imply that there is a constant C(Reν) such that

|Jν(r)| ≤C(Reν)e10 |Imν |2(1+ r)−12

whenever Reν > 0. This yields that if Rez > n−12

, then there is a constant C′(t) such

that the function Kz obeys the estimate

|Kz(y)| ≤C′( n2+Rez)e10 |Imz|2(1+ |y|)−n−(Rez− n−1

2 ) , (4.1.19)

and hence it lies in L1(Rn). Using identity (3.1.10), whenever Rez > n−12

, we define

for an integrable function f on Tn and x ∈ Tn the operator

BzR( f )(x) = ∑

ℓ∈Zn

mz(ℓR) f (ℓ)e2πiℓ·x = ( f ∗Lz,R)(x) ,

where Lz,R is a function whose sequence of Fourier coefficients is mz(ℓR)ℓ∈Zn .

But the function Lz,R can be precisely identified. By the Poisson summation for-

mula (Theorem 3.2.8), which applies since both Kz(x) and mz(x) are bounded by a

constant multiple of (1+ |x|)−n−δ for some δ > 0, we have

Lz,R(x) = ∑k∈Zn

mz(kR)e2πix·k = Rn ∑

ℓ∈Zn

Kz((x+ ℓ)R)

for all x ∈ Tn. We show that the family Lz,RR>0 is an approximate identity on Tn

when Rez > n−12

; on this see the related Exercise 3.1.3. Obviously, using (4.1.19) we

have that∫

Tn|Lz,R(x)|dx =

∫

Rn|Kz(y)|dy =C′′(n,Rez)e10 |Imz|2 < ∞ (4.1.20)

for some constant C′′(n,Rez), and also

∫

TnLz,R(x)dx =

∫

RnKz(y)dy = mz(0) = 1

for all R > 0 when Rez > n−12

. Moreover, for δ < 12

using (4.1.19) we have

∫

δ≤sup j |x j |≤ 12

∣∣Lz,R(x)∣∣dx≤ Cn,z

RRez− n−12

∫

δ≤sup j |x j |≤ 12

∑ℓ∈Zn

1

|x+ ℓ|n+Rez− n−12

dx→ 0 ,

thus the integral of Lz,R over [−1/2,1/2]n \ [−δ ,δ ]n tends to zero as R→ ∞.

Using Theorem 1.2.19, we obtain these conclusions for Rez > n−12

:

(a) For f ∈ L1(Tn), BzR( f ) converge to f in L1 as R→ ∞.

(b) For f continuous on Tn, BzR( f ) converge to f uniformly as R→ ∞.

We turn to the corresponding results for 1 n−12

=⇒ supR>0

∥∥BzR

∥∥L1(Tn)→L1(Tn)

= C′′(n,Rez)e10 |Imz|2 (4.1.21)

Rez = 0 =⇒ supR>0

∥∥BzR

∥∥L2(Tn)→L2(Tn)

= ‖mz‖L∞ = 1 . (4.1.22)

The family of operators f → BzR( f ) is of admissible growth for all Rez ≥ 0, since

for all measurable subsets A, B of Tn we have

∣∣∣∣∫

TnBz

R(χA)χB dx

∣∣∣∣=∣∣∣∣ ∑

k∈Zn

χA(k)mz(k)χB(k)

∣∣∣∣≤ ∑|k|≤R

1≤CnRn ,

thus condition (1.3.23) holds. Moreover, hypothesis (1.3.24) of Theorem 1.3.7 holds

in view of (4.1.21) and (4.1.22). Applying Theorem 1.3.7 (or rather Exercise 1.3.4 in

which the strip [0,1]×R is replaced by the more general strip [a,b]×R) we obtain

that when α = Rez > (n−1)| 1p− 1

2|, we have

supR>0

∥∥BαR∥∥

Lp(Tn)→Lp(Tn)< ∞ .

Finally, using Corollary 4.1.3, we deduce that BαR ( f )→ f in Lp(Tn) as R→ ∞ for

all f ∈ Lp(Tn).

The preceding result is sharp in the case p = 1 (Theorem 4.2.5). For this rea-

son, the number α = (n−1)/2 is referred to as the critical index of Bochner–Riesz

summability.

Exercises

4.1.1. If f ∈ C ∞(Tn), then show that DnN ∗ f and

Dn

N ∗ f converge to f uniformly

and in Lp for 1≤ p≤ ∞.

4.1.2. Prove that

‖P+‖L2(T1)→L2(T1) = ‖P−‖L2(T1)→L2(T1) = ‖W‖L2(T1)→L2(T1) = 1 ,

where W ( f ) = f is the conjugate function on the circle. Moreover, show that the

mappings f →W ( f )+ f (0) and f →W ( f )− f (0) are isometries on L2(T1).

4.1.3. Let −∞ ≤ a j < b j ≤ +∞ for 1 ≤ j ≤ n. Consider the rectangular projection

operator defined on C ∞(Tn) by

P( f )(x) = ∑a j≤m j≤b j

f (m)e2πi(m1x1+···+mnxn) .

Prove that when 1 < p < ∞, P extends to a bounded operator from Lp(Tn) to itself

with bounds independent of the a j,b j.[Hint: Express P in terms of the Riesz projection P+.

]

4.1.4. Let Pr(t) be the Poisson kernel on T1 as defined in Exercise 3.1.7. For 0 <r < 1, define the conjugate Poisson kernel Qr(t) on the circle by

Qr(t) =−i+∞

∑m=−∞

sgn (m)r|m|e2πimt .

(a) For 0 < r < 1, prove the identity

Qr(t) =2r sin(2πt)

1−2r cos(2πt)+ r2.

(b) Prove that f (t) = limr→1(Qr ∗ f )(t) whenever f is smooth. Conclude that if f is

real-valued, then so is f .

(c) Let f ∈ L1(T1). Prove that the function

z → (Pr ∗ f )(t)+ i(Qr ∗ f )(t)

is analytic in z = re2πit on the open unit disc z ∈ C : |z|< 1.(d) Let f ∈ L1(T1). Conclude that the functions z → (Pr ∗ f )(t) and z → (Qr ∗ f )(t)are conjugate harmonic functions of z = re2πit in the region |z| < 1. The term con-

jugate Poisson kernel stems from this property.

4.1.5. Let f be in.Λα(T

1) for some 0 < α < 1. Prove that the conjugate function f

is well defined and can be written as

f (x) = limε→0

∫

ε≤|t|≤1/2f (x− t)cot(πt)dt

=∫

|t|≤1/2

(f (x− t)− f (x)

)cot(πt)dt.

[Hint: Use part (b) of Exercise 4.1.4 and the fact that Qr has integral zero over the

circle to write ( f ∗Qr)(x) =(( f − f (x)) ∗Qr

)(x), allowing use of the Lebesgue

dominated convergence theorem.]

4.1.6. Suppose that f is a real-valued function on T1 with | f | ≤ 1 and 0≤ λ < π/2.

(a) Prove that ∫

T1eλ f (t) dt ≤ 1

cos(λ ).

(b) Conclude that for 0≤ λ < π/2 we have

∫

T1eλ | f (t)| dt ≤ 2

cos(λ ).

[Hint: Part (a): Consider the analytic function F(z) on the disk |z| < 1 defined by

F(z) = −i(Pr ∗ f )(θ)+ (Qr ∗ f )(θ), where z = re2πiθ . Then Re eλF(z) is harmonic

and its average over the circle |z| = r is equal to its value at the origin, which is

cos(λ f (0))≤ 1. Let r ↑ 1 and use that for z = e2πit on the circle we have Re eλF(z) ≥eλ f (t) cos(λ ).

]

4.1.7. Prove that for 0 < α < 1 there is a constant Cα such that

∥∥ f∥∥ .Λα (T1)

≤Cα

∥∥ f∥∥ .Λα (T1)

.

[Hint: Using Exercise 4.1.5, for |h| ≤ 1/10 write f (x+h)− f (x) as

∫

|t|≤5|h|

(f (x− t)− f (x+h)

)cot(π(t +h))dt

−∫

|t|≤5|h|

(f (x− t)− f (x)

)cot(πt)dt

+

∫

5|h|≤|t|≤1/2

(f (x− t)− f (x)

)(cot(π(t +h))− cot(πt)

)dt

+(

f (x)− f (x+h))∫

5|h|≤|t|≤1/2cot(π(t +h))dt .

You may use the fact that cot(πt) = 1πt

+ b(t), where b(t) is a bounded function

when |t| ≤ 1/2. The case |h| ≥ 1/10 is easy.]

4.2 Divergence Theorems 255

4.1.8. The beta function is defined in Appendix A.2. Derive the identity

tα =1

B(α−β ,β+1)

∫ t

0(t− s)α−β−1sβ ds

and show that the function KαR (x) = ∑|m|≤R

(1− |m|2

R2

)αe2πim·x satisfies (4.1.17).

[Hint: Take t = 1− |m|2

R2 and change variables s = r2−|m|2R2 in the displayed identity.

]

4.2 A. E. Divergence of Fourier Series and Bochner–Riesz means

We saw in Proposition 3.4.6 that the Fourier series of a continuous function may

diverge at a point. As expected, the situation can only get worse as the functions

get worse. In this section we present an example, due to A. N. Kolmogorov, of an

integrable function on T1 whose Fourier series diverges almost everywhere. We also

prove an analogous result for the Bochner–Riesz means at the critical index.

4.2.1 Divergence of Fourier Series of Integrable Functions

It is natural to start our investigation with the case n = 1. We begin with the follow-

ing important result:

Theorem 4.2.1. There exists an integrable function on the circle T1 whose Fourier

series diverges almost everywhere.

Proof. The proof of this theorem is a bit involved, and we need a sequence of lem-

mas, which we prove first.

Lemma 4.2.2. (Kronecker) Suppose that N ∈ Z+ and

x1,x2, . . . ,xN ,1

is a linearly independent set over the rationals. Then for any ε > 0 and any complex

numbers z1,z2, . . . ,zN with |z j|= 1, there exists an integer L ∈ Z such that

|e2πiLx j − z j|< ε for all 1≤ j ≤ N.

Proof. Suppose that the assertion claimed is false. Then there is an ε > 0 and com-

plex numbers z j = e2πiθ j , j = 1, . . . ,N, with 0≤ θ j < 1, such that

m(x1, . . . ,xN) : m ∈ Z∩B((θ1, . . . ,θN),ε

)= /0 ,

where B((θ1, . . . ,θN),ε

)denotes a neighborhood in TN of radius ε centered at the

point (θ1, . . . ,θN). Pick a smooth, nonzero, and nonnegative function f on TN sup-

ported in B((θ1, . . . ,θN),ε

). Then f (m(x1, . . . ,xN)) = 0 for all m ∈ Z, but

f (0) =

∫

TNf (y)dy > 0 . (4.2.1)

Set x = (x1, . . . ,xN). Then we have

0 =1

M

M−1

∑m=0

f (mx) =1

M

M−1

∑m=0

(∑

ℓ∈ZN

f (ℓ)e2πiℓ·mx

)

= ∑ℓ∈ZN

f (ℓ)

(1

M

M−1

∑m=0

e2πim(ℓ·x))

= f (0)+ ∑ℓ∈ZN\0

f (ℓ)

(1

M

e2πiM(ℓ·x)−1

e2πi(ℓ·x)−1

).

Note that e2πi(ℓ·x)−1 = 0 because ℓ ·x = ℓ1x1+ · · ·+ℓNxN /∈ Z, since by assumption

the set x1,x2, . . . ,xN ,1 is linearly independent over the rationals. Observe that

f (ℓ)1

M

e2πiM(ℓ·x)−1

e2πi(ℓ·x)−1= f (ℓ)

1

M

M−1

∑m=0

e2πim(ℓ·x)

tends to 0 as M → ∞ for every fixed ℓ ∈ ZN and is bounded uniformly in M by

| f (ℓ)|which satisfies∑ℓ∈ZN | f (ℓ)|<∞. Using the Lebesgue dominated convergence

theorem, we obtain that

0 = f (0)+ limM→∞

∑ℓ∈ZN\0

f (ℓ)

(1

M

e2πiM(ℓ·x)−1

e2πi(ℓ·x)−1

)

= f (0)+ ∑ℓ∈ZN\0

f (ℓ) limM→∞

(1

M

e2πiM(ℓ·x)−1

e2πi(ℓ·x)−1

)

= f (0)+0 ,

which contradicts (4.2.1). Therefore the claimed L exists.

Lemma 4.2.3. There exists a positive constant c > 0 such that given any integer

N ≥ 2 there exists a positive measure µN on T1 with µN(T1) = 1 such that

supL≥1

∣∣(µN ∗DL

)(x)|= sup

L≥1

∣∣∣∣L

∑k=−L

µN(k)e2πikx

∣∣∣∣≥ c logN (4.2.2)

for almost all x ∈ T1 (c is a fixed constant).

Proof. Given irrational real numbers x1, . . . ,xN such that the set x1, . . . ,xN ,1 is

linearly independent over the rationals, we define Q[x1, . . . ,xN ] to be the field exten-

sion of Q consisting of all linear combinations of the form q0 + q1x1 + · · ·+ qNxN ,

where q j are rational numbers. Obviously Q[x1, . . . ,xN ] is a countable set. Fix

N ≥ 100 and choose points x j as follows:

0 < x1 <1

N< x2 <

2

N< x3 <

3

N< · · ·< N−1

N< xN < 1 (4.2.3)

and such that x1 /∈ Q, x2 /∈ Q[x1], . . . , xN /∈ Q[x1, . . . ,xN−1]. Then obviously the set

x1, . . . ,xN ,1 is linearly independent over the rationals. Let

EN =

x ∈ [0,1] : x− x1, . . . ,x− xN ,1 is linearly independent over Q

and observe that every x in [0,1] \Q[x1, . . . ,xN ] belongs to EN . Indeed, if x /∈ EN ,

then there are rational numbers q j such that

q0 +q1(x− x1)+ · · ·+qN(x− xN) = 0.

Then q = q1 + · · ·+qN = 0, since x1, . . . ,xN ,1 are linearly independent over Q. It

follows that

x =−q−1q0 +q−1q1x1 + · · ·+q−1qNxN ,

thus x ∈Q[x1, . . . ,xN ]. We conclude that EN has full measure.

Next, we define the probability measure

µN =1

N

N

∑j=1

δx j,

where δx jare Dirac delta masses at the points x j. For this measure we have

∣∣∣∣L

∑k=−L

µN(k)e2πikx

∣∣∣∣ =

∣∣∣∣L

∑k=−L

(1

N

N

∑j=1

e−2πikx j

)e2πikx

∣∣∣∣

=

∣∣∣∣1

N

N

∑j=1

DL(x− x j)

∣∣∣∣

=

∣∣∣∣1

N

N

∑j=1

sin(2π(L+ 12)(x− x j))

sin(π(x− x j))

∣∣∣∣

=

∣∣∣∣1

N

N

∑j=1

Im[e2πi(L+ 1

2 )(x−x j)]sgn

(sin(π(x− x j))

)

|sin(π(x− x j))|

∣∣∣∣ ,

(4.2.4)

where the signum function is defined as sgna = 1 for a > 0, −1 for a < 0, and zero

if a = 0. By Lemma 4.2.2, for all x ∈ EN there exists an L ∈ Z+ such that

∣∣e2πiL(x−x j)− ie−2πi 12 (x−x j)sgn

(sin(π(x− x j))

)∣∣< 1

2,

which can be equivalently written as

∣∣e2πi(L+ 12 )(x−x j)sgn

(sin(π(x− x j))

)− i

∣∣< 1

2. (4.2.5)


Im[e2πi(L+ 1

2 )(x−x j)]sgn

(sin(π(x− x j))

)>

1

2.

Combining this with the result of the calculation in (4.2.4), we obtain that

∣∣∣∣L

∑k=−L

µN(k)e2πikx

∣∣∣∣>1

2N

N

∑j=1

1

|sin(π(x− x j))|≥ 1

2πN

N

∑j=1

1

|x− x j|.

But for every x ∈ [0,1), there exists a j0 such that x ∈ [x j0 ,x j0+1). It follows from

(4.2.3) that |x− x j| ≤C(| j− j0|+1)N−1, and thus

N

∑j=1

1

|x− x j|≥ c′N logN .

Thus for every x ∈ EN there exists an L ∈ Z+ such that

∣∣DL ∗µN(x)∣∣=

∣∣∣∣L

∑k=−L

µN(k)e2πikx

∣∣∣∣> c logN ,

which proves the required conclusion since EN is a set of measure 1.

Lemma 4.2.4. For each 0 < M <∞ there exists a trigonometric polynomial gM and

a measurable subset AM of T1 with measure |AM|> 1−2−M such that ‖gM‖L1 = 1,

and such that

infx∈AM

supL≥1

∣∣(DL ∗gM)(x)∣∣= inf

x∈AM

supL≥1

∣∣∣∣L

∑k=−L

gM(k)e2πikx

∣∣∣∣> 2M . (4.2.6)

Proof. Given an M ∈ Z+, we pick an integer N(M) such that c logN(M) > 2M+2,

where c is as in (4.2.2), and we also pick the measure µN(M), which satisfies (4.2.2).

By Fatou’s lemma we have

1 =∣∣x ∈ T1 : sup

L≥1

|(DL ∗µN(M))(x)| ≥ 2M+2∣∣

=∣∣ ⋃

L≥1

x ∈ T1 : sup

1≤ j≤L

|(D j ∗µN(M))(x)| ≥ 2M+2∣∣

=∫

T1limL→∞

χx∈T1: sup1≤ j≤L |(D j∗µN(M))(x)|≥2M+2dx

≤ liminfL→∞

∣∣x ∈ T1 : sup1≤ j≤L

|(D j ∗µN(M))(x)| ≥ 2M+2∣∣ ,

and thus we can find a positive integer L(M) such that the set

AM =

x ∈ T1 : sup1≤L≤L(M)

|(DL ∗µN(M))(x)| ≥ 2M+2

has measure greater than 1−2−M . We pick a positive integer K(M) such that

sup1≤ j≤L(M)

∥∥FK(M) ∗D j−D j

∥∥L∞≤ 1 ,

where FK is the Fejer kernel. This is possible, since the Fejer kernel is an approx-

imate identity and D j : 1 ≤ j ≤ L(M) is a finite family of continuous functions.

Then we define gM = µN(M)∗FK(M). Since µN(M) is a probability measure, we obtain

|(D j ∗gM)(x)− (D j ∗µN(M))(x)| ≤∥∥D j ∗FK(M)−D j

∥∥L∞≤ 1

for all x∈ [0,1] and 1≤ j≤ L(M). But given x∈AM there exists an L in 1, . . . ,L(M)such that |(DL ∗µN(M))(x)| ≥ 2M+2 and for this L we have

|(DL ∗gM)(x)| ≥ |(DL ∗µN(M))(x)|−1≥ 2M+2−1≥ 2M+1 > 2M .

Therefore, (4.2.6) is satisfied for this gM and AM . Since µN is a nonnegative measure

and FK(M) is nonnegative and has L1 norm 1, we have that

∥∥gM

∥∥L1 =

∥∥µN(M) ∗FK(M)

∥∥L1 =

∥∥µN(M)

∥∥M

∥∥FK(M)

∥∥L1 = 1 ,

showing that gM has L1 norm equal to one.

We now have the tools needed to construct an example of a function whose

Fourier series diverges almost everywhere. The example is given as a series of func-

tions whose behavior worsens as its index becomes bigger. The function we wish to

construct is a sum of the form

g =∞

∑j=1

ε jgM j, (4.2.7)

for a choice of sequences ε j → 0 and M j → ∞, where gM are as in Lemma 4.2.4.

Let us be specific. First, we set d0 = 1 and for N ≥ 1

dN = max1≤s≤N

degree(gMs), (4.2.8)

where gM is the trigonometric polynomial of Lemma 4.2.4. We set ε0 = M0 = 1.

Assume that we have defined ε j and M j for all 1≤ j < N for some N ≥ 2. We set

εN = 2−N(3dN−1)−1 (4.2.9)

and then we pick MN such that

2MN ≥(2N +dN−1 +1

)ε−1

N . (4.2.10)

This defines εN and MN for a given positive integer N, provided ε j and M j are known

for all j < N. This way we define εN and MN for all natural numbers N.

We observe that the selections of ε j and M j force the inequalities ε j ≤ 2− j and

d j ≤ d j+1 for all j ≥ 1. Since each gM jhas L1 norm 1 and ε j ≤ 2− j, the function g

in (4.2.7) is integrable and has L1 norm at most 1.

For a given j ≥ 1 and x ∈ AM j, by Lemma 4.2.4 there exists an L ≥ 1 such that

|(DL ∗gM j)(x)|> 2M j . Set

k = k(x) = min(L,d j).

Then we have

|(Dk ∗g)(x)| ≥ ε j|(Dk ∗gM j)(x)|− ∑

1≤s< j

εs|(Dk ∗gMs)(x)|−∑s> j

εs|(Dk ∗gMs)(x)| .

We make the following observations:

(i) |(Dk ∗gM j)(x)|= |(DL ∗gM j

)(x)|> 2M j .

(ii) |(Dk ∗gMs)(x)|= |(Dmin(ds,k) ∗gMs)(x)| ≤∥∥Dmin(ds,L)

∥∥L∞≤ 3ds, when s < j.

(iii) |(Dk ∗gMs)(x)|= |(Dmin(ds,k) ∗gMs)(x)| ≤∥∥Dmin(d j ,L)

∥∥L∞≤ 3d j, when s > j.

In these estimates we have used that k = min(L,d j),∥∥Dm

∥∥L∞≤ 2m+ 1 ≤ 3m, and

that

Dr ∗gMs = Dmin(r,ds) ∗gMs ,

which follows easily by examining the corresponding Fourier coefficients.

Using the estimates in (i), (ii), and (iii), for a fixed x∈ AM jand k = k(x) we obtain

|(Dk ∗g)(x)| ≥ ε j 2M j −3 ∑1≤s< j

εs ds−3∑s> j

εs d j . (4.2.11)

Our selection of ε j and M j now ensures that (4.2.11) is a large number. In fact, we

have

3∑s> j

εs d j = ∑s> j

2−sd j(ds−1)−1 ≤ ∑

s> j

2−s ≤ 1

and

3 ∑1≤s< j

εs ds ≤ 3d j−1 ∑1≤s< j

εs ≤ d j−1 ∑1≤s< j

2−s(ds−1)−1 ≤ d j−1 .


Therefore, the expression in (4.2.11) is at least ε j2M j −d j−1−1≥ 2 j. It follows that

for every j ≥ 1 and every x ∈ AM jthere exists a k = k(x) ∈ Z+ such that

|(Dk ∗g)(x)| ≥ 2 j .

We conclude that for every r ≥ 1 and x ∈∞⋃

j=r

AM j,

supk≥1

|(Dk ∗g)(x)| ≥ 2 j ≥ 2r , (4.2.12)

since x belongs to some AM jwith j ≥ r. For given r ≥ 1, Lemma 4.2.4 yields that

1≥∣∣∣∞⋃

j=r

AM j

∣∣∣≥ liminfj→∞

|AM j| ≥ lim

j→∞(1−2−M j) = 1.

Then the set

A =∞⋂

r=0

∞⋃

j=r

AM j

has measure 1, since it is a countable intersection of subsets of T1 of full measure.

In view of (4.2.12) we have that for all x in A

supk≥1

|(Dk ∗g)(x)| ≥ supr≥1

2r = ∞ (4.2.13)

and thus the required conclusion follows.

4.2.2 Divergence of Bochner–Riesz Means of Integrable Functions

We now turn to the corresponding n-dimensional problem for spherical summability

of Fourier series. The situation here is quite similar at the critical index α = n−12

.

Theorem 4.2.5. Let n > 1. There exists an integrable function f on Tn such that

limsupR→∞

∣∣∣Bn−1

2R ( f )(x)

∣∣∣= limsupR→∞

∣∣∣∣ ∑m∈Zn

|m|≤R

(1− |m|2

R2

) n−12 f (m)e2πim·x

∣∣∣∣= ∞

for almost all x ∈ Tn. Furthermore, such a function can be constructed such that it

is supported in an arbitrarily small given neighborhood of the origin.

We will need a couple of lemmas.

Lemma 4.2.6. Let n≥ 2. The complement of the set

S =

x ∈ Rn : 1∪|x−m| : m ∈ Zn is linearly independent over Q

has n-dimensional Lebesgue measure zero.

Proof. Recall that a function g defined on an open subset Ω of Rn is called real

analytic if for every point x0 in Ω there is a ball B(x0,ε) is contained in Ω and there

exist coefficients cβ (x0) such that g(x) = ∑β cβ (x0)(x− x0)β for all |x− x0| < ε ,

where the sum is taken over all multiindices. We will need two facts about real

analytic functions. First, the function x → |x| is real analytic on Rn \ 0. Indeed,

given x0 = 0, for |x− x0|< |x0|/3 we have that

|x− x0|2|x0|2

+2(x− x0) ·x0

|x0|2< 1 .

This allows us to write

|x|= |x0|(

1+|x− x0|2|x0|2

+2(x− x0) ·x0

|x0|2) 1

2

= |x0|∞

∑k=0

(1/2

k

)( |x− x0|2|x0|2

+2(x− x0) ·x0

|x0|2)k

,

which is a power series of the form ∑β cβ (x0)(x− x0)β .

Secondly, we need the fact that a real analytic function defined on an open con-

nected subset of Rn cannot vanish on a set of positive measure, unless it is identically

equal to zero; a proof of this in dimension one and an outline of the proof in higher

dimensions is contained in [205].

We return to the proof of the lemma which requires us to show that S has full

measure in Rn. Indeed, if x ∈ Rn \ S, then there exist k ∈ Z+, m1, . . .mk ∈ Zn, and

q0,q1, . . . ,qk nonzero rational numbers such that

q0 +k

∑j=1

q j|x−m j|= 0 . (4.2.14)

Since the function

y → q0 +k

∑j=1

q j|y−m j|

is nonzero and real analytic on Rn \Zn, it must vanish only on a set of Lebesgue

measure zero. Therefore, there exists a set Am1,...,mk,q0,q1,...,qkof Lebesgue measure

zero such that (4.2.14) holds exactly when x is in this set. Then

Rn \S⊆∞⋃

k=1

⋃

m1,...,mk∈Zn

⋃

q0,q1,...,qk∈Q

Am1,...,mk,q0,q1,...,qk,

from which it follows that Rn \S has Lebesgue measure zero.


Let us denote the Bochner–Riesz kernel by

KαR (x) = ∑

|m|≤R

(1− |m|2

R2

)αe2πim·x

when x ∈ Tn. We need the following lemma regarding KαR :

Lemma 4.2.7. Let n≥ 2. For almost every x ∈ Tn we have

limsupR→∞

|Kn−1

2R (x)|= ∞ .

It is noteworthy to compare the result of this lemma with the analogous one-

dimensional statement

limsupR→∞

|DR(x)|= ∞

for the Dirichlet kernel, which holds exactly when x = 0. Thus the uniform ill be-

havior of the kernel Kn−1

2R reflects in some sense its lack of localization.

Proof. Fix n ≥ 2 and fix x0 ∈([−1/2,1/2)n \ 0

)∩ S, where S is as in Lemma

4.2.6. Using (4.1.18) and the Poisson summation formula (Theorem 3.2.8), for each

α > n−12

we obtain the identity

KαR (x0) =

Γ (α+1)

παRn ∑

m∈Zn

J n2+α

(2πR|x0−m|)(R|x0−m|) n

2+α(4.2.15)

and the sum converges absolutely because of the asymptotics for the Bessel func-

tions in Appendix B.8. The term with m= 0 in the sum in (4.2.15) is a finite constant

since by Appendix B.6 the function

y →J n

2+α(2πR|y|)|y| n

2+α

is smooth and therefore bounded. But for m = 0 in (4.2.15) we have |m− x0| ≥ 1/2.

The asymptotics in Appendix B.8 imply that for R≥ 2 we have

J n2+α

(2πR|x0−m|) =e2πiR|x0−m|e−i π2 (

n2+α)−i π4 + e−2πiR|x0−m|ei π2 (

n2+α)+i π4

π√

R|x0−m|+O

((R|x0−m|)− 3

2)

for all α > n−12

. We insert this expression in (4.2.15), we multiply by e2πiλR for some

λ real, and then we average in R from 1 to T , for some T > 10. We obtain

1

T

∫ T

1Kα

R (x0)e2πiλR dR

=Γ (α+1)

πα ∑m∈Zn\0

e−i π2 (n2+α)−i π4

|x0−m| n+12 +α

1

T

∫ T

1e2πiR(λ+|x0−m|)R

n−12 −α dR

+Γ (α+1)

πα ∑m∈Zn\0

ei π2 (n2+α)+i π4

|x0−m| n+12 +α

1

T

∫ T

1e2πiR(λ−|x0−m|)R

n−12 −α dR

+Γ (α+1)

πα ∑m∈Zn\0

O

(1

|x0−m| n+32 +α

)1

T

∫ T

1R

n−32 −α dR

+Γ (α+1)

πα1

T

∫ T

1

J n2+α

(2πR|x0|)(R|x0|)

n2+α

e2πiλR Rn dR .

(4.2.16)

Assume that we are able to pass the limit as α → n−12+ through the sums and

integrals in the preceding identity; we justify this step momentarily. Then we obtain

1

T

∫ T

1K

n−12

R (x0)e2πiλR dR

=Γ ( n+1

2)

πn−1

2∑

m∈Zn\0

e−i π2 (2n−1

2 )−i π4

|x0−m|n1

T

∫ T

1e2πiR(λ+|x0−m|) dR

+Γ ( n+1

2)

πn−1

2∑

m∈Zn\0

ei π2 (2n−1

2 )+i π4

|x0−m|n1

T

∫ T

1e2πiR(λ−|x0−m|) dR

+Γ ( n+1

2)

πn−1

2∑

m∈Zn\0O

(1

|x0−m|n+1

)1

T

∫ T

1

dR

R

+Γ ( n+1

2)

πn−1

2

1

T

∫ T

1

Jn− 1

2(2πR|x0|)

(R|x0|)n− 12

e2πiλR Rn dR .

(4.2.17)

We now justify the passage of the limit in α inside the sums and the integrals in

(4.2.16) to obtain (4.2.17). First, when |m| ≤ R ≤ T and α > n−12

, the mean value

theorem gives

∣∣∣∣(

1− |m|2

R2

) n−12

e2πix0·m−(

1− |m|2

R2

)α

e2πix0·m∣∣∣∣

≤(α− n−1

2

)(1− |m|

2

R2

) n−12

log1

1− |m|2R2

≤(α− n−1

2

)sup

0<t≤1

tn−1

2 log1

t,

thus KαR (x0) converges to K

n−12

R (x0) uniformly in R∈ [1,T ] as α→ n−12

and therefore

the integral over [1,T ] of the former converges to the integral over [1,T ] of the latter.

Next, an integration by parts shows that the integral

∫ T

1e2πiR(λ+|x0−m|)R

n−12 −α dR

is bounded by a constant multiple of (λ + |x0−m|)−1, which makes the first infinite

sum in (4.2.16) converge absolutely and uniformly in α ≥ n−12

, thus one may pass

the limit in α inside the sum. Also, the integral

∫ T

1e2πiR(λ−|x0−m|)R

n−12 −α dR

is bounded by a constant multiple of (λ −|x0−m|)−1 whenever λ is not in the set

Λx0= |x0−m| : m ∈ Zn= λ1,λ2,λ3, . . . ,

where 0 < λ1 < λ2 < λ3 < · · · . Thus for λ /∈ Λx0, the preceding argument explains

the passage of the limit in α inside the second infinite sum in (4.2.16). If λ happens

to be in Λx0, then there is at most one m0 = 0, such that λ = |x0 −m0| and the

second sum in (4.2.16) restricted to m ∈ Zn \ 0,m0 converges absolutely, while

for the single term with m = m0, letting α → n−12+ is trivial. Finally, for the term

involving the Bessel function Jn− 1

2, the passage of the limit in α inside the integral

is straightforward since the function

(α,R) →J n

2+α(2πR|x0|)

(R|x0|)n2+α

is continuous on the compact set [ n−12, n

2]× [1,T ]. This completes the proof of

(4.2.17).

There are four terms to the right of (4.2.17) and we observe that if λ =±|x0−m0|for any m0 ∈ Zn, then all these terms converge to zero as T → ∞. This assertion

is trivial for the first three of these four terms, while for the last we assume that

T > |x0|−1. We split the integral

1

T

∫ T

1

Jn− 1

2(2πR|x0|)

(R|x0|)n− 12

e2πiλR Rn dR (4.2.18)

as a sum of the integral over [1, |x0|−1], which obviously converges to zero as T →∞by Appendix B.6, and of the integral over [ |x0|−1,T ]. For the latter, we use the

asymptotics in Appendix B.8 to write

Jn− 1

2(2πR|x0|) =

e2πiR|x0|e−i π2 (2n−1

2 )−i π4 + e−2πiR|x0|ei π2 (2n−1

2 )+i π4

π√

R|x0|+O

((R|x0|)−

32).

The part of the integral in (4.2.18) over [ |x0|−1,T ] corresponding to O((R|x0|)−

32

)

grows like log(T |x0|) which divided by T obviously tends to zero. The part of the

integral in (4.2.18) over [ |x0|−1,T ] corresponding to the main term is

1

T

1

π√|x0|

∫ T

|x0|−1e2πiR(λ+|x0|)e−i π2 (

2n−12 )−i π4 + e2πiR(λ−|x0|)ei π2 (

2n−12 )+i π4 dR

which tends to zero as T → ∞ by an integration by parts, since λ = |x0| because we

are considering the case where λ =±|x0−m| for any m ∈ Zn.

Now consider the case where λ =±|x0−m0| for some m0 ∈ Zn. In this case the

expression to the right in (4.2.17) converges to

Γ ( n+12)

πn−1

2

e±i( π2 (2n−1

2 )+ π4 )

|x0−m0|n=Γ ( n+1

2)

πn−1

2

e±i πn2

|x0−m0|n

as T → ∞. Next observe that∞

∑j=1

1

λ nj

= ∞ . (4.2.19)

We have now shown that

limT→∞

1

T

∫ T

1K

n−12

t (x0)e2πiλ t dt =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

Γ (n+1

2)

πn−1

2

ei πn

2

λ nj

if λ = λ j,

0 if λ =±λ j,

Γ (n+1

2)

πn−1

2

e−i πn

2

λ nj

if λ =−λ j.

(4.2.20)

Since x0 lies in S, the set 1 ∪ λ1,λ2,λ3, . . . is linearly independent over the

rationals and thus no expression of the form ±λ j1 ±·· ·±λ js is equal to an integer.

It follows from this fact and (4.2.20) that

limT→∞

1

T

∫ T

1K

n−12

t (x0)N

∏j=1

[1+

e−i πn2 e2πiλ jt + ei πn

2 e−2πiλ jt

2

]dt =

Γ ( n+12)

πn−1

2

N

∑j=1

1

λ nj

.

Suppose we had that

supR≥1

|Kn−1

2R (x0)| ≤ Ax0

< ∞ .

Then, setting cn =Γ (

n+12

)

πn−1

2

, we would have

cn

N

∑j=1

1

λ nj

= limT→∞

1

T

∫ T

1K

n−12

t (x0)N

∏j=1

[1+


2 e−2πiλ jt

2

]dt

= limsupT→∞

1

T

∫ T

1|K

n−12

t (x0)|N

∏j=1

∣∣∣∣1+e−i πn

2 e2πiλ jt + ei πn2 e−2πiλ jt

2

∣∣∣∣dt

≤ Ax0limsup

T→∞

1

T

∫ T

1

N

∏j=1

[1+


2 e−2πiλ jt

2

]dt

= Ax0,

which contradicts (4.2.19) by letting N → ∞. Here, once again, we used the fact

that no expression of the form ±λ j1 ± ·· ·±λ js is equal to an integer and thus the

preceding limsup is a limit and is equal to 1, since the integral of all the exponentials

produces another exponential which remains bounded.

We deduce that supR≥1 |Kn−1

2R (x0)| = ∞ for every point x0 ∈ S∩ [− 1

2, 1

2)n \ 0

and this concludes the proof of Lemma 4.2.7.

Proof. We now prove Theorem 4.2.5. This part of the proof is similar to the proof

of Theorem 4.2.1. Lemma 4.2.7 says that the means Bn−1

2R (δ0)(x), where δ0 is the

Dirac mass at 0, do not converge for almost all x ∈ Tn. Our goal is to replace this

Dirac mass by a series of integrable functions on Tn that have a peak at the origin.

Let us fix a nonnegative C ∞ radial function Φ on Rn that is supported in the unit

ball |ξ | ≤ 1 and has integral equal to 1. We set

ϕε(x) = ∑m∈Zn

1εn Φ( x+m

ε ) = ∑m∈Zn

Φ(εm)e2πim·x,

where the identity is valid because of the Poisson summation formula. It follows that

the mth Fourier coefficient of ϕε is Φ(εm). Therefore, we have the estimate

supx∈Tn

supR>0

|Bn−1

2R (ϕε)(x)| ≤ ∑

m∈Zn

|Φ(εm)| ≤ ∑m∈Zn

C′n(1+ε |m|)n+1

≤ Cn

εn. (4.2.21)

For any k ≥ 1, we construct measurable subsets Ek of Tn with |Ek| ≥ 1− 1k, a se-

quence of positive numbers R1 < R2 < · · · , with Rk ↑ ∞, and two sequences of posi-

tive numbers εk ↓ 0 and γk ↓ 0 such that εk ≤ γk for all k and

supR≤Rk

∣∣∣Bn−1

2R

( ∞

∑s=1

2−s(ϕεs −ϕγs))(x)

∣∣∣≥ k for x ∈ Ek. (4.2.22)

We pick E1 = /0, R1 = 1, and ε1 = γ1 = 1. Let k > 1 and suppose that we have selected

E j, R j, γ j, and ε j for all 1 ≤ j ≤ k− 1 such that (4.2.22) is satisfied. We construct

Ek, Rk, γk, and εk such that (4.2.22) is satisfied with j = k. We begin by choosing γk.

Let B be a constant such that

|Φ(x)−Φ(y)| ≤ B |x− y|

for all x,y ∈ Rn. Define γk such that

Bγk ∑|m|≤Rk−1

|m|= 1 . (4.2.23)

Then define

Ak =Cn2−kγ−nk +Cn

k−1

∑j=1

2− j(ε−nj + γ−n

j ) ,

where Cn is the constant in (4.2.21), and observe that in view of (4.2.21) we have

supx∈Tn

supR>0

∣∣∣Bn−1

2R

(−2−kϕγk

+k−1

∑j=1

2− j(ϕε j−ϕγ j

))(x)

∣∣∣≤ Ak . (4.2.24)

Let δ0 be the Dirac mass at the origin in Tn. Since by Fatou’s lemma and Lemma

4.2.7 we have

liminfR′→∞

∣∣∣

x ∈ Tn : sup0<R≤R′

∣∣∣Bn−1

2R (δ0)(x)

∣∣∣> 2k(Ak + k+2)∣∣∣= 1 ,

there exists an Rk > max(Rk−1,k) such that the set

Ek =

x ∈ Tn : sup0<R≤Rk

∣∣∣Bn−1

2R (2−kδ0)(x)

∣∣∣> Ak + k+2

has measure at least 1− 1k. Note that since Rk is increasing and tends to infinity,

(4.2.23) yields that γk is decreasing and tends to zero.

We now choose εk such that εk ≤ γk, εk ≤ εk−1, and that

supx∈Tn

supR≤Rk

2−k∣∣B

n−12

R (δ0)(x)−Bn−1

2R (ϕεk

)(x)∣∣≤ ∑

|m|≤Rk

2−k(1− |m|2

R2k

) n−12 |1−ϕεk

(m)| ≤ 1.

This is possible, since for a fixed Rk, the preceding sum tends to zero as εk → 0.

Then for x ∈ Ek we have

infx∈Ek

supR≤Rk

2−k∣∣B

n−12

R (ϕεk)(x)

∣∣≥ Ak + k+1 . (4.2.25)

The inductive selection of the parameters can be schematically described as follows:

γk−1,Rk−1,Ek−1,εk−1 =⇒ γk =⇒ Ak =⇒ Rk,Ek =⇒ εk =⇒ γk,Rk,Ek,εk.

Observe that the construction of γk gives for all s≥ k+1 the estimate

supx∈Tn

supR≤Rk

|Bn−1

2R (ϕεs −ϕγs)(x)| ≤ ∑

|m|≤Rk

|Φ(εsm)−Φ(γsm)|

≤ B(γs− εs) ∑|m|≤Rk

|m|

≤ Bγs ∑|m|≤Rk

|m|

≤ Bγk+1 ∑|m|≤Rk

|m|= 1 ,

(4.2.26)

using (4.2.23) and the fact that the sequence γk is decreasing.


We now prove (4.2.22). For x ∈ Tn write

Bn−1

2R

( ∞

∑s=1

2−s(ϕεs−ϕγs))(x) = B

n−12

R

(−2−kϕγk

+k−1

∑s=1

2−s(ϕεs−ϕγs))(x)

+Bn−1

2R

(2−kϕεk

)(x)

+Bn−1

2R

( ∞

∑s=k+1

2−s(ϕεs−ϕγs))(x) ,


supR≤Rk

∣∣∣∣Bn−1

2R

( ∞

∑s=1


∣∣∣∣≥ supR≤Rk

∣∣∣∣Bn−1

2R

(2−kϕεk

)(x)

∣∣∣∣

− supR≤Rk

∣∣∣∣Bn−1

2R

(−2−kϕγk

+k−1

∑s=1


∣∣∣∣

− supR≤Rk

∣∣∣∣Bn−1

2R

( ∞

∑s=k+1


∣∣∣∣ .

In view of (4.2.25), (4.2.24), and (4.2.26) for all x ∈ Ek, we obtain

supR≤Rk

∣∣∣Bn−1

2R

( ∞

∑s=1

2−s(ϕεs −ϕγs))(x)

∣∣∣≥ (Ak + k+1)−Ak−∞

∑s=k+1

2−s ≥ k ,

which clearly implies (4.2.22). Setting

f =∞

∑s=1

2−s(ϕεs −ϕγs) ∈ L1(Tn) ,

we deduce that supR>0

∣∣Bn−1

2R ( f )(x)

∣∣≥ k for all x in⋃∞

r=k Er, and thus

supR>0

∣∣Bn−1

2R ( f )(x)

∣∣= ∞

for all x in∞⋂

k=1

∞⋃

r=k

Er .

Since this set has full measure in Tn, the required conclusion follows.

By taking ε1 arbitrarily small (instead of picking ε1 = 1), we force f to be sup-

ported in an arbitrarily small neighborhood of the origin.

The previous argument shows that the Bochner–Riesz means BαR are badly be-

haved on L1(Tn) when α = n−12

. It follows that the “rougher” spherical Dirichlet

means

DnN ∗ f (which correspond to α = 0) are also ill behaved on L1(Tn). See

Exercise 4.2.2.

Exercises

4.2.1. Using Theorem 4.2.1 construct a function F on Tn such that

limsupN→∞

|(DnN ∗F)(x1, . . . ,xn)|= ∞

for almost all (x1, . . . ,xn) ∈ Tn.

4.2.2. For any 0≤ α < ∞ and R > 0 consider the Bochner–Riesz kernel

KαR (x) = ∑

|m|≤R

(1− |m|2

R2

)αe2πim·x .

Use Exercise 4.1.8 to obtain that if for some x0 ∈ Tn we have

limsupR→∞

|KαR (x0)|< ∞ ,

then for all β > α we have

supR>0

|KβR (x0)|< ∞ .

Conclude that whenever 0 ≤ α ≤ n−12

, the Bochner–Riesz means of order α of

the function f constructed in the proof of Theorem 4.2.5, in particular the circu-

lar (spherical) Dirichlet means of this function, diverge a.e.

4.2.3. (a) Show that for M,N positive integers we have

(FM ∗DN)(x) =

⎧⎨⎩

FM(x) for M ≤ N,

FN(x)+M−N

(M+1)(N+1) ∑|k|≤N

|k|e2πikx for M > N.

(b) Prove that for some constant c > 0 we have

∫

T1

∣∣∣∣ ∑|k|≤N

|k|e2πikx

∣∣∣∣dx≥ cN logN

as N → ∞.[Hint: Part (b): Show that for x ∈ [− 1

2, 1

2] we have

∑|k|≤N

|k|e2πikx = (N +1)(DN(x)−FN(x))

and use the result of Exercise 3.1.5.]

4.3 Multipliers, Transference, and A. E. Convergence 271

4.2.4. Given the integrable functions

f1(x) =∞

∑j=0

2− jF222 j (x) , f2(x) =

∞

∑j=1

1

j2F

22 j (x) , x ∈ T1,

show that ‖ f1 ∗DN‖L1 → ∞ and ‖ f2 ∗DN‖L1 → ∞ as N → ∞.[Hint: Let M j = 222 j

or M j = 22 jdepending on the situation. For fixed N let jN

be the least integer j such that M j > N. Then for j ≥ jN + 1 we have M j ≥M2jN>

N2 ≥ 2N+1, henceM j−N

M j+1≥ 1

2. Split the summation indices into the sets j≥ jN and

j < jN . Conclude that ‖ f1 ∗DN‖L1 and ‖ f2 ∗DN‖L1 tend to infinity as N → ∞ using

Exercise 4.2.3.]

4.3 Multipliers, Transference, and Almost Everywhere

Convergence

In Chapter 2 we saw that bounded operators from Lp(Rn) to Lq(Rn) that commute

with translations are given by convolution with tempered distributions on Rn. In par-

ticular, when p = q, these tempered distributions have bounded Fourier transforms,

called Fourier multipliers. Convolution operators that commute with translations

can also be defined on the torus. These lead to Fourier multipliers on the torus.

4.3.1 Multipliers on the Torus

In analogy with the nonperiodic case, we could identify convolution operators on Tn

with appropriate distributions on the torus; see Exercise 4.3.2 for an introduction to

this topic. However, it is simpler to avoid this point of view and consider the study

of multipliers directly, bypassing the discussion of distributions on the torus.

For h ∈ Tn we define the translation operator τh acting on a periodic function

f as follows: τh( f )(x) = f (x− h) for x ∈ Tn. We say that a linear operator T act-

ing on functions on the torus commutes with translations if for all h ∈ Tn we have

τh(T ( f ))(x) = T (τh f )(x) for almost all x ∈ Tn.

Theorem 4.3.1. Suppose that T is a linear operator that commutes with translations

and maps Lp(Tn) to Lq(Tn) for some 1 ≤ p,q ≤ ∞. Then there exists a bounded

sequence amm∈Zn such that

T ( f )(x) = ∑m∈Zn

am f (m)e2πim·x (4.3.1)

for all f ∈ C ∞(Tn). Moreover, we have

∥∥am∥∥ℓ∞≤

∥∥T∥∥

Lp→Lq .

Proof. Consider the functions em(x) = e2πim·x defined on Tn for m in Zn. Since

T commutes with translations, for every h ∈ Tn there is a subset Fh of Tn of full

measure such that

T (em)(x−h) = T (τh(em))(x) = e−2πim·hT (em)(x)

for every x ∈ Fh. Note that

∫

Tn|h ∈ Tn : x ∈ Fh|dx =

∫

Tn

∫

Tnχ(h,x)∈Tn×Tn: x∈Fh dhdx

=∫

Tn

∫

Tnχ(h,x)∈Tn×Tn: x∈Fh dxdh

=∫

Tn|Fh|dh = 1.

Therefore there exists an x0 ∈ Tn such that |h ∈ Tn : x0 ∈ Fh|= 1. It follows that

for almost all h ∈ Tn we have T (em)(x0−h) = e−2πim·hT (em)(x0). Replacing x0−h

by x, we obtain

T (em)(x) = e2πim·x(e−2πim·x0T (em)(x0))= amem(x) (4.3.2)

for almost all x ∈ Tn, where we set am = e−2πim·x0T (em)(x0), for m ∈ Zn. Taking Lq

norms in (4.3.2), we deduce |am|= ‖T (em)‖Lq ≤‖T‖Lp→Lq , and thus am is bounded.

Moreover, since T (em) = amem for all m in Zn, it follows that (4.3.1) holds for all

trigonometric polynomials. By density this extends to all f ∈ C ∞(Tn) and the theo-

rem is proved.

Definition 4.3.2. Let 1 ≤ p,q ≤ ∞. We call a bounded sequence amm∈Zn an

(Lp,Lq) multiplier if the corresponding operator given by (4.3.1) maps Lp(Tn)to Lq(Tn). If p = q, (Lp,Lp) multipliers are simply called Lp multipliers. When

1 ≤ p < ∞, the space of all Lp multipliers on Tn is denoted by Mp(Zn). This no-

tation follows the convention that Mp(G) denote the space of Lp multipliers on

Lp(G), where G is a locally compact group and G is its dual group. The norm of

an element am in Mp(Zn) is the norm of the operator T given by (4.3.1) from

Lp(Tn) to itself. This norm is denoted by∥∥am

∥∥Mp

.

We now examine some special cases. We begin with the case p = q = 2. As

expected, it turns out that M2(Zn) = ℓ∞(Zn).

Theorem 4.3.3. A linear operator T that commutes with translations maps L2(Tn)to itself if and only if there exists a sequence amm∈Zn in ℓ∞(Zn) such that

T ( f )(x) = ∑m∈Zn

am f (m)e2πim·x (4.3.3)

for all f ∈ C ∞(Tn). Moreover, in this case we have∥∥T

∥∥L2→L2 =

∥∥amm

∥∥ℓ∞

.


Proof. The existence of such a sequence is guaranteed by Theorem 4.3.1, which also

gives ‖amm‖ℓ∞ ≤ ‖T‖L2→L2 . Conversely, any operator of the form (4.3.3) satisfies

∥∥T ( f )∥∥2

L2 = ∑m∈Zn

|am f (m)|2 ≤∥∥amm

∥∥2

ℓ∞ ∑m∈Zn

| f (m)|2 ,

and thus ‖T‖L2→L2 ≤ ‖amm‖ℓ∞ .

We continue with the case p = q = 1. Recall the definition of a finite Borel mea-

sure on Tn. Given such a measure µ , its Fourier coefficients are defined by

µ(m) =

∫

Tne−2πix·m dµ(x), m ∈ Zn .

Clearly, all the Fourier coefficients of the measure µ are bounded by the total vari-

ation∥∥µ

∥∥ of µ . See Exercise 4.3.3 for basic properties of Fourier transforms of

distributions on the torus.

Theorem 4.3.4. A linear operator T that commutes with translations maps L1(Tn)to itself if and only if there exists a finite Borel measure µ on the torus such that

T ( f )(x) = ∑m∈Zn

µ(m) f (m)e2πim·x (4.3.4)

for all f ∈C ∞(Tn). Moreover, in this case we have ‖T‖L1→L1 = ‖µ‖. In other words,

M1(Zn) is the set of all sequences given by Fourier coefficients of finite Borel mea-

sures on Tn.

Proof. Fix f ∈ L1(Tn). If (4.3.4) is valid, then T ( f )(m) = f (m)µ(m) for all m∈Zn.

But Exercise 4.3.3 gives that f ∗µ(m) = f (m)µ(m) for all m ∈ Zn; therefore, the

integrable functions f ∗ µ and T ( f ) have the same Fourier coefficients, and they

must be equal. Thus T ( f ) = f ∗ µ , which implies that T is bounded on L1 and

‖T ( f )‖L1 ≤ ‖µ‖‖ f‖L1 .

To prove the converse direction, we suppose that T commutes with translations

and maps L1(Tn) to itself. We recall the Poisson kernel Pε defined on Tn, which can

be expressed in the following two ways:

Pε(x) = ∑m∈Zn

e−2π|m|εe2πim·x =Γ ( n+1

2)

πn+1

2∑

m∈Zn

ε−n

(1+ | x+mε |2)

n+12

≥ 0 (4.3.5)

for all x ∈Tn, in view of the identity obtained in (3.2.4). The preceding identity says

that Pε ≥ 0; hence, ‖Pε‖L1 =∫

Tn Pε(x)dx. Integrating the first series in (4.3.5) over

Tn we conclude that ‖Pε‖L1(Tn) = 1. The boundedness of T now gives

∥∥T (Pε)∥∥

L1(Tn)≤

∥∥T∥∥

L1→L1

for all ε > 0. The Banach–Alaoglu theorem implies that there exist a sequence ε j ↓ 0

and a finite Borel measure µ on Tn such that T (Pε j) tends to µ weakly as j → ∞.

This means that for all continuous functions g on Tn we have

limj→∞

∫

Tng(x)T (Pε j

)(x)dx =∫

Tng(x)dµ(x) . (4.3.6)

It follows from (4.3.6) that for all g continuous on Tn we have

∣∣∣∣∫

Tng(x)dµ(x)

∣∣∣∣≤ supj

∥∥T (Pε j)∥∥

L1

∥∥g∥∥

L∞≤

∥∥T∥∥

L1→L1

∥∥g∥∥

L∞.

Since, by the Riesz representation theorem we have that the norm of the linear func-

tional

g →∫

Tng(x)dµ(x)

on the space of continuous functions C (Tn) is ‖µ‖, it follows that

‖µ‖ ≤∥∥T

∥∥L1→L1 . (4.3.7)

It remains to prove that T has the form given in (4.3.4). By Theorem 4.3.1 we have

that there exists a bounded sequence am on Zn such that (4.3.1) is satisfied. Taking

g(x) = e−2πik·x in (4.3.6) and using the representation for T in (4.3.1), we obtain

µ(k) =∫

Tne−2πik·x dµ(x)

= limj→∞

∫

Tne−2πik·x ∑

m∈Zn

ame−2πε j |m|e2πim·x dx

= limj→∞

∑m∈Zn

∫

Tne−2πik·xame−2πε j |m|e2πim·x dx

= limj→∞

ake−2πε j |k| = ak .

This proves assertion (4.3.4). It follows from (4.3.4) that T ( f ) = f ∗ µ and thus

‖T‖L1→L1 ≤ ‖µ‖. This fact combined with (4.3.7) gives ‖T‖L1→L1 = ‖µ‖.

Remark 4.3.5. It is not hard to see that most basic properties of the space Mp(Rn)

of Lp Fourier multipliers on Rn are also valid for Mp(Zn). In particular, Mp(Z

n) is

a closed subspace of ℓ∞(Zn) and thus a Banach space itself. Moreover, sums, scalar

multiples, and products of elements of Mp(Zn) are also in Mp(Z

n), which makes

this space a Banach algebra. As in the nonperiodic case, we also have Mp(Zn) =

Mp′(Zn) when 1 < p < ∞.

4.3.2 Transference of Multipliers

It is clear by now that multipliers on L1(Tn) and L1(Rn) are very similar, and the

same is true for L2(Tn) and L2(Rn). These similarities became obvious when we

characterized L1 and L2 multipliers on both Rn and Tn. So far, it is not known if a

nontrivial characterization of Mp(Rn) exists, but we might ask whether this space is

related to Mp(Zn). There are several connections of this type and there are general

ways to produce multipliers on the torus from multipliers on Rn and vice versa.

General methods of this sort are called transference of multipliers.

We begin with a useful definition.

Definition 4.3.6. Let t0 ∈Rn. A bounded function b on Rn is called regulated at the

point t0 if

limε→0

1

εn

∫

|t|≤ε

(b(t0− t)−b(t0)

)dt = 0 . (4.3.8)

The function b is called regulated if it is regulated at every t0 ∈ Rn.

Clearly, if t0 is a Lebesgue point of b, then b is regulated at t0. In particular, this

is the case if b is continuous at t0. If b(t0) = 0, condition (4.3.8) also holds when

b(t0− t) = −b(t0 + t) whenever |t| ≤ ε for some ε > 0; for instance the function

b(t) =−isgn(t− t0) has this property.

An example of a regulated function is the following modification of the charac-

teristic function of the cube [−1,1]n

χ[−1,1]n(x1, . . . ,xn) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

1 when all |x j|< 1,

2k−n if (x1, . . . ,xn) belongs to some k-dimensional

face of the boundary of [−1,1]n,

0 when some |x j|> 1,

with the understanding that points are zero-dimensional.

The first transference result we discuss is the following.

Theorem 4.3.7. Suppose that b is a regulated function at every point m ∈ Zn and

that b lies in Mp(Rn) for some 1 0 the sequences b(m/R)m∈Zn are

in Mp(Zn) and we have

supR>0

∥∥b(m/R)m∈Zn

∥∥Mp(Zn)

≤ ‖b‖Mp(Rn) .

The second conclusion of the theorem is a consequence of the first, since for a

given R > 0 the function b(ξ/R) is regulated on Zn and has the same Mp(Rn) norm

as b(ξ ). Before we prove this result, we state and prove a couple of lemmas.

Lemma 4.3.8. Suppose that the function b on Rn is regulated at the point x0. Let

Kε(x) = ε−ne−π|x/ε |2

for ε > 0. Then we have that (b∗Kε)(x0)→ b(x0) as ε → 0.

Proof. For r > 0 define the function

Fx0(r)=

1

rn

∫

|t|≤r

(b(x0−t)−b(x0)

)dt =

1

rn

∫ r

0

∫

Sn−1

(b(x0−sθ)−b(x0)

)dθsn−1ds.

Let η > 0. Since b is regulated at x0 there is a δ > 0 such that for r ≤ δ we have

|Fx0(r)| ≤ η . Fix such a δ and write

(b∗Kε)(x0)−b(x0) =∫

y∈Rn

(b(x0− y)−b(x0)

)Kε(y)dy = Aε1 +Aε2,

where

Aε1 =∫

|y|≥δ

(b(x0− y)−b(x0)

)Kε(y)dy

and

Aε2 =

∫

|y|<δ

(b(x0− y)−b(x0))Kε(y)

)dy

=

∫ δ

0

1

εne−π(r/ε)

2∫

Sn−1

(b(x0− rθ)−b(x0)

)rndr

=∫ δ

0

1

εne−π(r/ε)

2 d

dr(rnFx0

(r))dr .

For our given η > 0 there is an ε0 > 0 such that for ε < ε0 we have

|Aε1| ≤ 2‖b‖L∞

∫

|y|≥ δε

e−π|y|2

dy < η .

Via an integration by parts Aε2 can be written as

|Aε2| =∣∣∣∣δ nFx0

(δ )1

εne−π(δ/ε)

2 −0+2π∫ δ

0

r

εn+2e−π(r/ε)

2

rnFx0(r)dr

∣∣∣∣

=∣∣∣Fx0

(δ )δ n

εne−π(δ/ε)

2

+2π∫ δ/ε

0rn+1Fx0

(εr)e−πr2

dr

∣∣∣

≤∣∣Fx0

(δ )∣∣δ

n

εne−π(δ/ε)

2

+ sup0<r≤ δ

ε

|Fx0(εr)|2π

∫ δ/ε

0rn+1e−πr2

dr

≤∣∣Fx0

(δ )∣∣Cn + sup

0<r≤δ|Fx0

(r)|C′n

≤ (Cn +C′n)η ,

where we set Cn = supt>0 tne−πt2and C′n = 2π

∫ ∞0 rn+1e−πr2

dr. Then for ε < ε0 we

have |(b∗Kε)(x0−b(x0)|< (Cn +C′n +1)η , thus (b∗Kε)(x0)→ b(x0) as ε → 0.


Lemma 4.3.9. Let T be the operator on Rn whose multiplier is b(ξ ), and let S be

the operator on Tn whose multiplier is the sequence b(m)m∈Zn . Assume that b(ξ )is regulated at every point ξ = m ∈ Zn. Suppose that P and Q are trigonometric

polynomials on Tn and let Lε(x) = e−πε |x|2

for x ∈Rn and ε > 0. Then the following

identity is valid whenever α,β > 0 and α+β = 1:

limε→0

εn2

∫

RnT (PLεα)(x)Q(x)Lεβ (x)dx =

∫

TnS(P)(x)Q(x)dx. (4.3.9)

Proof. It suffices to prove the required assertion for P(x) = e2πim·x and Q(x) =e2πik·x, k,m ∈ Zn, since the general case follows from this case by linearity. In view

of Parseval’s relation (Proposition 3.2.7 (3)), we have

∫

TnS(P)(x)Q(x)dx = ∑

r∈Zn

b(r)P(r)Q(r) =

b(m) when k = m,

0 when k = m.(4.3.10)

On the other hand, using the identity in Theorem 2.2.14 (3), we obtain

εn2

∫

RnT (PLεα)(x)Q(x)Lεβ (x)dx

= εn2

∫

Rnb(ξ )PLεα(ξ )QLεβ (ξ )dξ

= εn2

∫

Rnb(ξ )(εα)−

n2 e−π

|ξ−m|2εα (εβ )−

n2 e−π |ξ−k|2

εβ dξ

= (εαβ )−n2

∫

Rnb(ξ )e−π

|ξ−m|2εα e

−π |ξ−k|2εβ dξ . (4.3.11)

Now if m = k, since α+β = 1, the expression in (4.3.11) is equal to

(εαβ )−n2

∫

Rnb(ξ )e

−π |ξ−m|2εαβ dξ , (4.3.12)

which tends to b(m) in view of Lemma 4.3.8, since b is regulated at every point

m ∈ Zn.

We now consider the case m = k in (4.3.11). Since |m− k| ≥ 1, then every ξ in

Rn must satisfy either |ξ −m| ≥ 1/2 or |ξ − k| ≥ 1/2. Therefore, the expression in

(4.3.11) is controlled by

(εαβ )−n2

(∫

|ξ−m|≥ 12

b(ξ )e−π

4εα e−π |ξ−k|2

εβ dξ +∫

|ξ−k|≥ 12

b(ξ )e− π

4εβ e−π|ξ−m|2εα dξ

),

which is in turn controlled by

‖b‖L∞(α−

n2 e−

π4εα +β−

n2 e− π

4εβ),


which tends to zero as ε → 0. This proves that the expression in (4.3.10) is equal

to the limit of the expression in (4.3.11) as ε → 0. This completes the proof of

Lemma 4.3.9

Proof (Theorem 4.3.7). We are assuming that T maps Lp(Rn) to itself and we need

to show that S maps Lp(Tn) to itself. We prove this using duality. For P and Q

trigonometric polynomials, using Lemma 4.3.9, we have

∣∣∣∣∫

TnS(P)(x)Q(x)dx

∣∣∣∣

=

∣∣∣∣ limε→0

εn2

∫

RnT (PLε/p)(x)Q(x)Lε/p′(x)dx

∣∣∣∣

≤∥∥T

∥∥Lp→Lp limsup

ε→0

εn2

∥∥PLε/p

∥∥Lp(Rn)

∥∥QLε/p′∥∥

Lp′(Rn)

=∥∥T

∥∥Lp→Lp limsup

ε→0

(ε

n2

∫

Rn|P(x)|pe−επ|x|

2

dx

)1p(ε

n2

∫

Rn|Q(x)|p′e−επ|x|2 dx

) 1p′

=∥∥T

∥∥Lp→Lp

(∫

Tn|P(x)|p dx

)1p(∫

Tn|Q(x)|p′ dx

) 1p′,

provided for all continuous 1-periodic functions g on Rn we have that

limε→0

εn2

∫

Rng(x)e−επ|x|

2

dx =∫

Tng(x)dx. (4.3.13)

Assuming (4.3.13) for the moment, we take the supremum over all trigonometric

polynomials Q on Tn with Lp′ norm at most 1 to obtain that S maps Lp(Tn) to itself

with norm at most ‖T‖Lp→Lp , yielding the required conclusion.

We now prove (4.3.13). Use the Poisson summation formula to write the left-hand

side of (4.3.13) as

εn2 ∑

k∈Zn

∫

Tng(x− k)e−επ|x−k|2 dx =

∫

Tng(x)ε

n2 ∑

k∈Zn

e−επ|x−k|2 dx

=∫

Tng(x) ∑

k∈Zn

e−π|k|2/εe2πix·k dx

=

∫

Tng(x)dx+Aε ,

where

|Aε | ≤∥∥g

∥∥L∞ ∑|k|≥1

e−π|k|2/ε → 0

as ε → 0. This completes the proof of Theorem 4.3.7.

We now obtain a converse of Theorem 4.3.7. If b(ξ ) is a bounded function on Rn

and the sequence b(m)m∈Zn is in Mp(Zn), then we cannot necessarily obtain that

b is in Mp(Rn), since such a conclusion would depend on the values of b on the in-

teger lattice, which is a set of measure zero. However, a converse can be formulated

if we assume that for all R > 0, the sequences b(m/R)m∈Zn are in Mp(Zn) uni-

formly in R. Then we obtain that b(ξ/R) is in Mp(Rn) uniformly in R > 0, which is

equivalent to saying that b∈Mp(Rn), since dilations of multipliers on Rn do not af-

fect their norms (see Proposition 2.5.14). These remarks can be precisely expressed

in the following theorem.

Theorem 4.3.10. Suppose that b(ξ ) is a bounded function defined on Rn which is

Riemann integrable over any cube. Suppose that the sequences b(mR)m∈Zn are in

Mp(Zn) uniformly in R > 0 for some 1 0

∥∥b(mR)m∈Zn

∥∥Mp(Zn)

. (4.3.14)

Proof. Suppose that f and g are smooth functions with compact support on Rn.

Then there is an R0 > 0 such that for R≥ R0, the functions x → f (Rx) and x → g(Rx)are supported in [−1/2,1/2]n. We define periodic functions

FR(x) = ∑k∈Zn

f (R(x− k)) and GR(x) = ∑k∈Zn

g(R(x− k))

on Tn. Observe that the mth Fourier coefficient of FR is FR(m) = R−n f (m/R) and

that of GR is GR(m) = R−ng(m/R).Now for R≥ R0 we have

∣∣∣∣ ∑m∈Zn

b(m/R) f (m/R)g(m/R) Volume(

mR+[0, 1

R]n)∣∣∣∣ (4.3.15)

=

∣∣∣∣Rn ∑m∈Zn

b(m/R)FR(m)GR(m)

∣∣∣∣

=

∣∣∣∣Rn

∫

Tn

(∑

m∈Zn

b(m/R)FR(m)e2πim·x)

GR(x)dx

∣∣∣∣

≤ Rn∥∥b(m/R)m

∥∥Mp(Zn)

∥∥FR

∥∥Lp(Tn)

∥∥GR

∥∥Lp′ (Tn)

≤ supR>0

∥∥b(m/R)m∈Zn

∥∥Mp(Zn)

Rn∥∥FR

∥∥Lp(Rn)

∥∥GR

∥∥Lp′ (Rn)

= supR>0

∥∥b(m/R)m∈Zn

∥∥Mp(Zn)

∥∥ f∥∥

Lp(Rn)

∥∥g∥∥

Lp′ (Rn). (4.3.16)

Since b is bounded and Riemann integrable over any cube in Rn, the function

b(ξ ) f (ξ )g(ξ ) is Riemann integrable over Rn. The expressions in (4.3.15) are sums

associated with the partition [mR, m+1

R)nm∈Zn of Rn which tend to

∣∣∣∣∫

Rnb(ξ ) f (ξ )g(ξ )dξ

∣∣∣∣

as R → ∞ by the definition of the Riemann integral. We deduce that the absolute

value of ∫

Rnb(ξ ) f (ξ )g(ξ )dξ =

∫

Rn(b f )∨(x)g(x)dx

is bounded by the expression in (4.3.16). This proves the theorem via duality.

4.3.3 Applications of Transference

Having established two main transference theorems, we turn to an application.

Corollary 4.3.11. Let 1 < p < ∞, f ∈ Lp(Tn), and α ≥ 0. Then

(a)∥∥Dn

R ∗ f − f∥∥

Lp(Tn)→ 0 as R→ ∞ if and only if χ[−1,1]n ∈Mp(R

n).

(b)∥∥ Dn

R ∗ f − f∥∥

Lp(Tn)→ 0 as R→ ∞ if and only if χB(0,1) ∈Mp(R

n).

(c)∥∥BαR ( f )− f

∥∥Lp(Tn)

→ 0 as R→ ∞ if and only if (1−|ξ |2)α+ ∈Mp(Rn).

Proof. First observe that in view of Corollary 4.1.3, the assertions on the left in (a),

(b), and (c) are equivalent to the statements

supR>0

∥∥DnR ∗ f

∥∥Lp(Tn)

≤Cp

∥∥ f∥∥

Lp(Tn),

supR>0

∥∥ DnR ∗ f

∥∥Lp(Tn)

≤Cp

∥∥ f∥∥

Lp(Tn),

supR>0

∥∥BαR ( f )∥∥

Lp(Tn)≤Cp

∥∥ f∥∥

Lp(Tn),

for some constant 0 < Cp < ∞ and all f in Lp(Tn). These statements can be

rephrased as

supR>0

∥∥χ[−1,1]n(m/R)m∈Zn

∥∥Mp(Zn)

<∞ ,

supR>0

∥∥χB(0,1)(m/R)m∈Zn

∥∥Mp(Zn)

<∞ ,

supR>0

∥∥(1−|m/R|2)α+m∈Zn

∥∥Mp(Zn)

<∞ .

If these statements hold, then Theorem 4.3.10 gives that the functions χ[−1,1]n(ξ ),

χB(0,1)(ξ ), and (1−|ξ |2)α+ lie in Mp(Rn).

To prove the converse implication, for any given R′ ∈ R+ \ |m| : m ∈ Zn, the

functions χ[−1,1]n(ξ/R′), χB(0,1)(ξ/R′) are Riemann integrable over Rn and are reg-

ulated (actually continuous) at every point in Zn. Moreover, the function (1−|ξ |2)α+is continuous, regulated, and Riemann integrable over Rn. Then the hypotheses of

Theorem 4.3.7 are satisfied and its conclusion yields that

∥∥χ[−1,1]n(m/R′)m∈Zn

∥∥Mp(Zn)

≤∥∥χ[−1,1]n

∥∥Mp

, (4.3.17)


∥∥χB(0,1)(m/R′)m∈Zn

∥∥Mp(Zn)

≤∥∥χB(0,1)

∥∥Mp

, (4.3.18)

supR>0

∥∥(1−|m/R|2)α+m∈Zn

∥∥Mp(Zn)

≤∥∥(1−| · |2)α+

∥∥Mp

.

Notice that the first and second estimates are uniform in R′, so one may insert a

supremum over R′ ∈R+ \|m| : m ∈ Zn in (4.3.17) and (4.3.18). To replace R′ by a

general R∈Z+ simply notice that for any R > 0 there is an R′ ∈R+ \|m| : m∈Znsuch that

DnR ∗ f = Dn

R′ ∗ f andDn

R ∗ f =

DnR′ ∗ f

for any f ∈ Lp(Tn). Then using (4.3.17) we obtain

supR>0

∥∥DnR ∗ f

∥∥Lp = sup

R>0

∥∥DnR′ ∗ f

∥∥Lp = sup

R′>0

∥∥DnR′ ∗ f

∥∥Lp ≤

∥∥χ[−1,1]n∥∥

Mp

∥∥ f∥∥

Lp

and likewise forDn

R.

4.3.4 Transference of Maximal Multipliers

We now prove a theorem concerning maximal multipliers analogous to Theorems

4.3.7 and 4.3.10. This enables us to reduce problems related to almost everywhere

convergence of Fourier series on the torus to problems of boundedness of maximal

operators on Rn.

Let b be a bounded function defined on all of Rn. For R > 0, we introduce the

multiplier operators

Sb,R(F)(x) = ∑m∈Zn

b(m/R)F(m)e2πim·x , (4.3.19)

Tb,R( f )(x) =∫

Rnb(ξ/R) f (ξ )e2πiξ ·x dξ , (4.3.20)

initially defined for smooth functions with compact support f on Rn and smooth

functions F on Tn.

We introduce the maximal operators

Mb(F)(x) = supR>0

∣∣Sb,R(F)(x)∣∣ , (4.3.21)

Nb( f )(x) = supR>0

∣∣Tb,R( f )(x)∣∣ , (4.3.22)

defined for smooth functions F on Tn and smooth functions with compact support

f on Rn. Let τy(b)(ξ ) = b(ξ − y) be a translation operator defined for y ∈ Rn. We

have the following result concerning these operators.

Theorem 4.3.12. Let b be a function defined on Rn. Suppose that b is bounded,

regulated, Riemann integrable over any cube, and assume that for all ξ ∈ Rn the

function t → b(ξ/t) has only countably many discontinuities on R+. Let 1 < p < ∞and Cp < ∞, and suppose that b lies in Mp(R

n). Let Mb and Nb be as in (4.3.21)

and (4.3.22). Then the following assertions are equivalent:

∥∥Mb(F)∥∥

Lp(Tn)≤Cp

∥∥b∥∥

Mp

∥∥F∥∥

Lp(Tn), F ∈ C

∞(Tn), (4.3.23)∥∥Nb( f )

∥∥Lp(Rn)

≤Cp

∥∥b∥∥

Mp

∥∥ f∥∥

Lp(Rn), f ∈ C

∞0 (Rn). (4.3.24)

Proof. Let F = t1, . . . , tk be a finite subset of R+. We prove the claimed equiva-

lences for the maximal operators

MFb (G)(x) = sup

t∈F

∣∣Sb,t(G)(x)∣∣ ,

NFb (g)(x) = sup

t∈F

∣∣Tb,t(g)(x)∣∣ ,

with constants that are uniform in the finite set F . Then MFb may be viewed as

an operator defined on the dense subspace C ∞(Tn) of Lp(Tn) and taking values in

Lp(Tn, ℓ∞(F )), which is the dual space of Lp′(Tn, ℓ1(F )). Likewise, NFb is defined

on the dense subspace C ∞0 (Rn) of Lp(Rn) and takes values in Lp(Rn, ℓ∞(F )), which

is the dual space of Lp′(Rn, ℓ1(F )). Using duality, with respect to the complex inner

product, estimates (4.3.23) and (4.3.24) are equivalent to the pair of inequalities

∣∣∣∣ ∑m∈Zn

G(m)k

∑j=1

b(m

t j

)Fj(m)

∣∣∣∣≤Cp

∥∥b∥∥

Mp

∥∥G∥∥

Lp(Tn)

∥∥∥k

∑j=1

|Fj|∥∥∥

Lp′ (Tn), (4.3.25)

∣∣∣∣∫

Rng(ξ )

k

∑j=1

b(ξ

t j

)f j(ξ )dξ

∣∣∣∣≤Cp

∥∥b∥∥

Mp

∥∥g∥∥

Lp(Rn)

∥∥∥k

∑j=1

| f j|∥∥∥

Lp′ (Rn), (4.3.26)

where g, f j ∈ C ∞0 (Rn), and G,Fj ∈ C ∞(Tn). In proving the equivalence of (4.3.25)

and (4.3.26), by density, we work with smooth functions with compact support g, f j

and trigonometric polynomials G,Fj.

Suppose that (4.3.25) holds and let f1, . . . , fk,g be smooth functions with com-

pact support on Rn. Then there is an R0 > 0 such that for R ≥ R0 the functions

Fj,R(x) = f j(Rx) and GR(x) = g(Rx) are supported in [−1/2,1/2]n and thus they

can be viewed as functions on Tn once they are periodized. Also, the mth Fourier

coefficient of Fj,R is Fj,R(m) = R−n f j(m/R) and that of GR is GR(m) = R−ng(m/R).Since b lies in Mp(R

n) we have

‖b‖Mp(Rn) = supR>0

∥∥b(m/R)m∈Zn

∥∥Mp(Zn)

in view of Theorems 4.3.7 and 4.3.10, which are both applicable in view of the

hypotheses of b.


As in the proof of Theorem 4.3.10, for R≥ R0 we have

∣∣∣∣ ∑m∈Zn

k

∑j=1

b(m/Rt j) f j(m/R)g(m/R)Volume(

mR+[0, 1

R]n)∣∣∣∣ (4.3.27)

=

∣∣∣∣Rn ∑m∈Zn

k

∑j=1

b(m/Rt j)Fj,R(m)GR(m)

∣∣∣∣

≤Cp‖b‖MpRn

∥∥∥k

∑j=1

|Fj,R|∥∥∥

Lp′ (Tn)

∥∥GR

∥∥Lp(Tn)

=Cp‖b‖Mp

∥∥∥k

∑j=1

| f j|∥∥∥

Lp′ (Rn)‖g‖Lp(Rn) ,

where we applied (4.3.25) in the first inequality above for the function ξ → b(ξ/R),which has the same Mp norm as b.

Since b is bounded and Riemann integrable over any cube in Rn, the functions

b(ξ/t j) f j(ξ )g j(ξ ) are Riemann integrable over Rn. Realizing the limit of the partial

sums in (4.3.27) when R→ ∞ as a Riemann integral, we obtain

∣∣∣∣∫

Rn

k

∑j=1

b(ξ/t j) f j(ξ )g(ξ )dξ

∣∣∣∣≤Cp

∥∥b∥∥

Mp

∥∥∥k

∑j=1

| f j|∥∥∥

Lp′ (Rn)‖g‖Lp(Rn)

and thus we showed that (4.3.25) implies (4.3.26).

We now turn to the converse. Assume that (4.3.26) holds. We will prove (4.3.25)

for trigonometric polynomials and then by density we extend it to all C ∞ functions

on Tn. Expressing g in terms of g in (4.3.26) and taking the supremum in (4.3.26)

over all C ∞0 functions g with Lp norm 1 we deduce that

∥∥∥∥∫

Rn

k

∑j=1

b(ξ

t j

)f j(ξ )e

2πi(·)·ξ dξ

∥∥∥∥Lp′≤ Cp

∥∥b∥∥

Mp

∥∥∥k

∑j=1

| f j|∥∥∥

Lp′ (Rn). (4.3.28)

Let P1, . . . ,Pk and Q be trigonometric polynomials on Tn. Set Lε(x) = e−πε |x|2.

Since b is regulated at every point in Rn, Lemma 4.3.9 gives

∣∣∣∣ ∑m∈Zn

k

∑j=1

Q(m)b(m/t j)Pj(m)

∣∣∣∣

=

∣∣∣∣∫

Tn

(∑

m∈Zn

k

∑j=1

Pj(m)b(m/t j)e2πim·x

)Q(x)dx

∣∣∣∣

=

∣∣∣∣ limε→0

εn2

∫

Rn

(∫

Rn

k

∑j=1

PjLε/p′(ξ )b(ξ/t j)e2πiξ ·xdξ

)Q(x)Lε/p(x)dx

∣∣∣∣


≤Cp‖b‖Mplimsupε→0

[ε

n2

∥∥∥k

∑j=1

∣∣PjLε/p′∣∣∥∥∥

Lp′ (Rn)

∥∥QLε/p

∥∥Lp(Rn)

]

=Cp‖b‖Mplimsupε→0

[ε

n2p′

∥∥∥( k

∑j=1

∣∣Pj

∣∣)

Lε/p′

∥∥∥Lp′ (Rn)

(ε

n2

∫

Rn|Q(x)|pe−επ|x|

2

dx

)1p]

=Cp‖b‖Mp

∥∥∥k

∑j=1

∣∣Pj

∣∣∥∥∥

Lp′ (Rn)

∥∥Q∥∥

Lp(Tn),

where we used Holder’s inequality and (4.3.28) in the only inequality above and

(4.3.13) in the last equality. Thus we obtain that (4.3.26) implies (4.3.25), and this

completes the equivalence of boundedness of MFb and NF

b .

We now prove the claimed equivalence for the operators Mb and Nb. We first

show that if MFb is bounded on (C ∞(Tn),‖ · ‖Lp) with bound independent of the

finite set F , then Mb is bounded on (C ∞(Tn),‖ · ‖Lp).For each ξ ∈ Rn, let Aξ be the null subset of R+ such that t → b(ξ/t) is contin-

uous on R+ \Aξ . We fix a function F in C ∞(Tn), and we note that for each x ∈ Tn

the function

t → Sb,t(F)(x) = ∑m∈Zn

b(m/t)F(m)e2πim·x (4.3.29)

is continuous on the set R+ \⋃m∈Zn Am. We pick a countable dense subset D′ of

R+ \⋃m∈Zn Am, and we let D = D′ ∪⋃m∈Zn Am. Then D is a countable set and the

Lebesgue monotone convergence theorem gives that

∥∥supt∈D

|Sb,t(F)|∥∥

Lp(Tn)= lim

k→∞

∥∥MFk

b (F)∥∥

Lp(Tn)≤Cp‖b‖Mp

∥∥F∥∥

Lp(Tn), (4.3.30)

where Fk is an increasing sequence of finite sets whose union is D. Using that the

function in (4.3.29) is continuous on R+ \D, we conclude that the supremum over

t ∈ D in (4.3.30) can be replaced by the supremum over t ∈ Z+ (Exercise 4.3.7).

Assume now that NFb is bounded on (C ∞

0 (Rn),‖ · ‖Lp) with bound independent

of the finite set F . We show that Nb is bounded on (C ∞0 (Rn),‖ · ‖Lp). Let f be in

C ∞0 (Rn). We have that the map

t → Tb,t( f )(x) =∫

Rnb(ξ/t) f (ξ )e2πiξ ·x dξ = tn

∫

Rnb(ξ ) f (tξ )e2πiξ ·tx dξ (4.3.31)

is a continuous function on R+ since f is continuous. Thus the estimate

∥∥supt∈D

|Tb,t( f )|∥∥

Lp(Rn)≤Cp‖b‖Mp

∥∥ f∥∥

Lp(Rn)(4.3.32)

for a countable dense subset D of R+ (such as D = Q+) can be easily extended by

replacing the supremum over D by the supremum over R+. And estimate (4.3.32)

for D = Q+ follows from the corresponding estimate on finite sets via the Lebesgue

monotone convergence theorem.

Remark 4.3.13. Under the hypotheses of Theorem 4.3.12, the following two in-

equalities are also equivalent:

∥∥Mb(G)∥∥

Lp,∞(Tn)≤Cp‖b‖Mp

∥∥G∥∥

Lp(Tn), G ∈ C

∞(Tn), (4.3.33)∥∥Nb(g)

∥∥Lp,∞(Rn)

≤C′p‖b‖Mp‖g‖Lp(Rn) , g ∈ C

∞0 (Rn), (4.3.34)

with C′p ≤Cp ≤C(n, p)C′p for some other constant C(n, p). Indeed, Exercise 1.4.12

gives that the pair of inequalities (4.3.33) and (4.3.34) is equivalent to the pair of

inequalities

∣∣∣∣ ∑m∈Zn

k

∑j=1

Fj(m)b(m/t j)G(m)

∣∣∣∣≤Cp‖b‖Mp

∥∥G∥∥

Lp(Tn)

∥∥∥k

∑j=1

|Fj|∥∥∥

Lp′ ,1(Tn), (4.3.35)

∣∣∣∣∫

Rn

k

∑j=1

f j(ξ )b(ξ/t j)g(ξ )dξ

∣∣∣∣≤C′p‖b‖Mp‖g‖Lp(Rn)

∥∥∥k

∑j=1

| f j|∥∥∥

Lp′ ,1(Rn), (4.3.36)

where Lp′,1 is the Lorentz space and f j ∈ C ∞0 (Rn) and Fj ∈ C ∞(Tn).

Now (4.3.36) follows from (4.3.35) just like (4.3.26) follows from (4.3.25) with

the only exception being that Holder’s inequality for Lp and Lp′ is replaced by

Holder’s inequality for Lp,∞ and Lp′,1 and we use that ‖g‖Lp,∞ ≤ ‖g‖Lp . Conversely,

assuming (4.3.36), in order to prove (4.3.35) it will suffice to know that

sup0<ε<1

εn2q

∥∥∥( k

∑j=1

∣∣Pj

∣∣)

Lε/q

∥∥∥Lq,1(Rn)

≤C(n,q)∥∥∥

k

∑j=1

∣∣Pj

∣∣∥∥∥

Lq,1(Tn). (4.3.37)

For this we refer to Exercise 4.3.6.

4.3.5 Applications to Almost Everywhere Convergence

As an application of the preceding results, we relate the almost everywhere conver-

gence of Fourier series of functions on T1 with the almost everywhere convergence

of Fourier integrals of functions on R. In this subsection we show that the following

two results are equivalent:

Theorem 4.3.14. For every 1 < p < ∞ there exists a finite constant Cp such that for

all F ∈ C ∞(T1) we have

∥∥∥ supN∈Z+

|F ∗DN |∥∥∥

Lp≤Cp

∥∥F∥∥

Lp . (4.3.38)

Theorem 4.3.15. For every 1 < p < ∞ there exists a finite constant Cp such that for

all f ∈ C ∞0 (R) we have

∥∥C∗∗( f )∥∥

Lp(R)≤Cp‖ f‖Lp(R) (4.3.39)

where

C∗∗( f )(x) = supR>0

∣∣∣∣∫

|ξ |≤Rf (ξ )e2πixξ dξ

∣∣∣∣

is the Carleson operator.

As a consequence of Theorem 4.3.14, we obtain that for any F ∈ Lp(T1), we have

limN→∞

∑|m|≤N

F(m)e2πimx = F(x)

for almost every x ∈ [0,1].Theorem 4.3.14 can be proved directly, but we do not pursue this here. Instead,

we show the equivalence of the two theorems and refer the interested reader to [131],

which contains the proof of Theorem 4.3.15.

We observe that both operators F → C∗(F) = supN>0 |F ∗DN | and f → C∗∗( f )are sublinear and take nonnegative values. Thus they satisfy the inequalities

|C∗(F)−C∗(G)| ≤ C∗(F−G) |C∗∗( f )−C∗∗(g)| ≤ C∗∗( f −g)

for all F,G in C ∞(T1) and f ,g in C ∞0 (R). Then, by density (see the argument in

the proof of Theorem 1.4.19 or Exercise 1.4.17), they admit bounded extensions to

Lp(T1) and Lp(R), respectively, so that (4.3.38) and (4.3.39) hold for all F ∈ Lp(T1)and f ∈ Lp(R).

Next, we discuss the details of the transference argument that claims the equiva-

lence of Theorems 4.3.14 and 4.3.15.

Consider the following function defined on R:

b(x) =

⎧⎪⎨⎪⎩

1 when |x|< 1,

1/2 when |x|= 1,

0 when |x|> 1.

(4.3.40)

Then b is bounded and Riemann integrable over any interval, and is easily seen to

be regulated; also, given any x ∈ R, the function t → b(x/t) is discontinuous only

for t ∈ x,−x.Let Sb,R be as in (4.3.19), where b is defined in (4.3.40). We note that inequality

(4.3.38) is equivalent to

∥∥∥supR>0

|Sb,R(F)|∥∥∥

Lp≤C′p

∥∥F∥∥

Lp (4.3.41)

for all F ∈ C ∞(T1), where DRR>0 is the family of Dirichlet kernels as defined in

(3.1.16), depending on the continuous parameter R. Indeed, we have

Sb,R(F)(x) =

⎧⎪⎪⎨⎪⎪⎩

∑|m|≤[R] F(m)e2πimx if R /∈ Z+,

(DR−1 ∗F)(x)+F(R)e2πixR+F(−R)e−2πixR

2if R ∈ Z+.

(4.3.42)

Since supR>0 |F(±R)| ≤ ‖F‖L1 ≤ ‖F‖Lp , it follows that if (4.3.38) holds, then

(4.3.41) also holds with C′p =Cp +1.

The only hypothesis of Theorem 4.3.12 missing is that b lies in Mp(R). We

obtain this from the fact that supR>0 ‖b(·/R)‖Mp(Z) < ∞ via Theorem 4.3.10, since

supR>0

∥∥F ∗DR

∥∥Lp(T1)

= supN∈Z+

∥∥F ∗DN

∥∥Lp(T1)

≤C′′p∥∥F‖Lp(T1) , (4.3.43)

where the last estimate follows from Proposition 4.1.6, Theorem 4.1.7, and Corollary

4.1.3. The preceding equality is due to the fact that DR = DR+ε whenever 0 < ε < 1.

Now all hypotheses of Theorem 4.3.12 are valid. As a consequence we obtain the

equivalence of the boundedness of the the maximal operator

Nb( f )(x) = C∗∗( f )(x) = supR>0

∣∣∣∣∫ +R

−Rf (ξ )e2πixξ dξ

∣∣∣∣

on Lp(R) and of

Mb(F)(x) = supR>0

∣∣∣∣ ∑m∈Z

F(m)e2πimxb(m

R

)∣∣∣∣= supR>0

|Sb,R(F)(x)| ,

on Lp(T1). But in view of (4.3.42) and of the fact that supR>0 |F(±R)| ≤ ‖F‖Lp , the

Lp boundedness of Mb is equivalent to the Lp boundedness of C∗ on Lp(T1). This

discussion concludes the equivalence of Theorems 4.3.14 and 4.3.15.

4.3.6 Almost Everywhere Convergence of Square Dirichlet Means

The extension of Theorem 4.3.14 to higher dimensions is a rather straightforward

consequence of the one-dimensional result.

Theorem 4.3.16. For every 1 0

|DnN ∗ f |

∥∥∥Lp(Tn)

≤Cp,n

∥∥ f∥∥

Lp(Tn)(4.3.44)

and consequently

limN→∞

∑m∈Zn

|m j |≤N

f (m)e2πim·x = f (x)

for almost every x ∈ Tn and f ∈ Lp(Tn).

Proof. We prove Theorem 4.3.16 when n = 2. Fix a p with 1 < p < ∞. Since the

Riesz projection P+ is bounded on Lp(T1) (Theorem 4.1.7 and identity (4.1.9)), ap-

plying Theorem 4.3.10 with b(ξ ) = χ(0,∞), we obtain that that the function χ(0,∞)is in Mp(R). It follows that the characteristic function of the half-space ξ1 > 0 in

R2 lies in Mp(R2). Since rotations and translations of multipliers preserve their

Mp norms (Proposition 2.5.14), it follows that the characteristic function of any half

space created by a line in R2 lies in Mp(R2) with a fixed norm. The product of three

multipliers is a multiplier (Proposition 2.5.13); thus the characteristic function of the

triangle T created by the lines ξ2 = ξ1− 14, ξ2 = −ξ1− 1

4, ξ2 = L+ 1

4lies also in

Mp(R2) with norm independent of L ∈ Z+. The regulated function

σ(ξ1,ξ2) =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

1 if (ξ1,ξ2) ∈ T \∂T

0 if (ξ1,ξ2) /∈ T12

if (ξ1,ξ2) ∈ ∂T \(0,− 14),(L+ 1

2,L+ 1

4),(−L− 1

2,L+ 1

4)

18

if (ξ1,ξ2) ∈ (L+ 12,L+ 1

4),(−L− 1

2,L+ 1

4)

14

if (ξ1,ξ2) = (0,− 14)

is a.e. equal to the characteristic function of T . Thus Theorem 4.3.7 gives that the

restriction of σ on Z2, i.e., the sequence am1,m2m1,m2

defined by am1,m2= 1 when

|m1| ≤ |m2| ≤ L and zero otherwise, lies in Mp(Z2) with norm independent of L in

Z+. This means that for some constant Bp we have the following inequality for all f

in Lp(T2):

∫

T2

∣∣∣∣ ∑m2∈Z|m2|≤L

∑m1∈Z

|m1|≤|m2|

f (m1,m2)e2πi(m1x1+m2x2)

∣∣∣∣p

dx2 dx1 ≤ Bpp

∥∥ f∥∥p

Lp(T2),

(4.3.45)

where Bp is independent of L ∈ Z+. There is also a a version of (4.3.45), proved

similarly, in which |m1| ≤ |m2| is replaced by the strict inequality |m1|< |m2|.Now let 1 < p < ∞, L ∈ Z+, and f ∈ Lp(T2). For fixed x1 ∈ T1 define

f Lx1(x2) = ∑

m2∈Z|m2|≤L

[∑

m1∈Z|m1|≤|m2|

f (m1,m2)e2πim1x1

]e2πim2x2 = ∑

m2∈Z|m2|≤L

f Lx1(m2)e

2πim2x2

and for fixed x2 ∈ T1 define

fx2L (x1) = ∑

m1∈Z|m1|≤L

[∑

m2∈Z|m2|<|m1|

f (m1,m2)e2πim2x2

]e2πim1x1 = ∑

m1∈Z|m1|≤L

fx2L (m1)e

2πim1x1 .

We have

∫

T1

∫

T1sup

0<N≤L

∣∣∣∣ ∑|m1|≤N

∑|m2|≤N

f (m1,m2)e2πim1x1e2πim2x2

∣∣∣∣p

dx2 dx1

≤ 2p−1∫

T1

∫

T1sup

0<N≤L

∣∣∣∣ ∑|m2|≤N

[∑

|m1|≤|m2|f (m1,m2)e

2πim1x1

]e2πim2x2

∣∣∣∣p

+ sup0<N≤L

∣∣∣∣ ∑|m1|≤N

[∑

|m2|<|m1|f (m1,m2)e

2πim2x2

]e2πim1x1

∣∣∣∣p

dx1 dx2

= 2p−1

[∫

T1

∫

T1sup

0<N≤L

∣∣(DN ∗ f Lx1)(x2)

∣∣pdx2 dx1

+∫

T1

∫

T1sup

0<N≤L

∣∣(DN ∗ fx2L )(x1)

∣∣pdx1 dx2

]

≤ 2p−1

[∫

T1

∫

T1sup

N∈Z+

∣∣(DN ∗ f Lx1)(x2)

∣∣pdx2dx1

+∫

T1

∫

T1sup

N∈Z+

∣∣(DN ∗ fx2L )(x1)

∣∣pdx1 dx2

]

≤ 2p−1Cpp

∫

T1

∫

T1| f L

x1(x2)|p dx2 dx1 +2p−1Cp

p

∫

T1

∫

T1| f x2

L (x1)|p dx1 dx2

≤ 2pCppBp

p

∥∥ f∥∥p

Lp(T2),

where we used Theorem 4.3.14 in the penultimate inequality and estimate (4.3.45)

in the last inequality. Since the last estimate we obtained is independent of L ∈ Z+,

letting L → ∞ and applying Fatou’s lemma, we obtain the conclusion (4.3.44) for

n = 2. When n ≥ 3 the idea of the proof is similar, but the notation a bit more

cumbersome.

Exercises

4.3.1. Let α ≥ 0. Prove that the function (1−|ξ |2)α+ is in Mp(Rn) if and only if the

function (1−|ξ |)α+ is in Mp(Rn).[

Hint: Use that smooth functions with compact support lie in Mp(Rn).

]

4.3.2. The purpose of this exercise is to introduce distributions on the torus. The

set of test functions on the torus is C ∞(Tn) equipped with the following topology.

Given f j, f in C ∞(Tn), we say that f j → f in C ∞(Tn) if

∥∥∂α f j−∂α f∥∥

L∞(Tn)→ 0 as j→ ∞, ∀ α .

Under this notion of convergence, C ∞(Tn) is a topological vector space with topol-

ogy induced by the family of seminorms ρα(ϕ) = supx∈Tn |(∂α f )(x)|, where α


ranges over all multi-indices. The dual space of C ∞(Tn) under this topology is the

set of all distributions on Tn and is denoted by D ′(Tn). The definition implies that

for u j and u in D ′(Tn) we have u j → u in D ′(Tn) if and only if

⟨u j, f

⟩→

⟨u, f

⟩as j→ ∞ for all f ∈ C

∞(Tn).

The following operations can be defined on elements of D ′(Tn): differentiation (as

in Definition 2.3.6), translation and reflection (as in Definition 2.3.11), convolution

with a C ∞ function (as in Definition 2.3.13), multiplication by a C ∞ function (as

in Definition 2.3.15), the support of a distribution (as in Definition 2.3.16). Use the

same ideas as in Rn to prove the following:

(a) Prove that if u ∈ D ′(Tn) and f ∈ C ∞(Tn), then ( f ∗u)(x) = 〈u,τx( f )〉 is a C ∞

function.

(b) In contrast to Rn, the convolution of two distributions on Tn can be defined. For

u,v ∈D ′(Tn) and f ∈ C ∞(Tn) define

⟨u∗ v, f

⟩=

⟨u, f ∗ v

⟩.

Check that convolution of distributions on D ′(Tn) is associative, commutative, and

distributive.

(c) Prove the analogue of Proposition 2.3.23, i.e., that C ∞(Tn) is dense in D ′(Tn).

4.3.3. For u ∈D ′(Tn) and m ∈ Zn define the Fourier coefficient u(m) by

u(m) = u(e−2πim·( ·)) =⟨u,e−2πim·( ·)⟩ .

Prove properties (1), (2), (4), (5), (6), (8), (9), (11), and (12) of Proposition 2.3.22

regarding the Fourier coefficients of distributions on the circle. Moreover, prove that

for any u, v in D ′(Tn) we have (u ∗ v)(m) = u(m) v(m). In particular, this is valid

for finite Borel measures.

4.3.4. Let µ be a finite Borel measure on Rn and let ν be the periodization of µ ,

that is, ν is a measure on Tn defined by

ν(A) = ∑m∈Zn

µ(A+m)

for all measurable subsets A of Tn. Prove that the restriction of the Fourier transform

of µ on Zn coincides with the sequence of the Fourier coefficients of the measure ν .

4.3.5. Let vn be the volume of the unit ball in Rn and e1 = (1,0, . . . ,0). Prove that

limε→0

1

vnεn

∫

|x−e1|≤εχ|x|≤1 dx =

1

2.

Conclude that the function

χB(0,1)(x) =

⎧⎪⎨⎪⎩

1 when |x|< 1,

1/2 when |x|= 1,

0 when |x|> 1

is regulated.

4.3.6. Let Lε(x) = e−πε |x|2

be defined for ε > 0 and x ∈ Rn and let 1 < q < ∞.

Prove that there is a constant C(n,q) < ∞ such that for any 1-periodic continuous

function g on Rn we have

sup0<ε<1

εn2q∥∥gLε/q

∥∥Lq,1(Rn)

≤C(n,q)‖g‖Lq,1(Tn) .

[Hint: Reduce matters to the situation where g = ∑k∈Zn χk+E , where E is a measur-

able subset of [−1/2,1/2)n. Express the Lq,1 norm of gLε/q in terms of its distribu-

tion function and for 0 < λ < 1 estimate the measure

∣∣Lε/q > λ⋂ ⋃

k∈Zn

(k+E)∣∣=

∣∣∣B(

0,( q

πεlog

1

λ

) 12)⋂ ⋃

k∈Zn

(k+E)∣∣∣

by Cn

(√n+

(qπε log 1

λ

)1/2)n |E|.

]

4.3.7. Let 0 < C0 < ∞. Suppose that ftt∈R+ is a family of measurable functions

on a measure space X that satisfies

∥∥supt∈F

| ft |∥∥

Lp ≤C0

for every finite subset F of R+.

(a) Suppose that for each x ∈ X , the function t → ft(x) is continuous. Show that

∥∥supt>0

| ft |∥∥

Lp ≤C0 .

(b) Prove that for any t > 0 there is a measurable function ft on X that is a.e. equal

to ft such that ∥∥ supt∈R+

| ft |∥∥

Lp ≤C0 .

[Hint: Part (a): Notice that in view of the Lebesgue monotone convergence theo-

rem, we have∥∥supt∈Q | ft |

∥∥Lp ≤ C0. Also, for each x ∈ X we have supt∈Q | ft(x)| =

supt∈R+ | ft(x)| by continuity. Part (b): Let a = supF ‖supt∈F | ft |‖Lp ≤C0, where the

supremum is taken over all finite subsets F of R. Pick an increasing sequence of

finite sets Fn such that ‖supt∈Fn| ft |‖Lp → a as n→ ∞. Let g = supn supt∈Fn

| ft | and

note that ‖g‖Lp = a. Then for any s ∈ R we have


∥∥∥max(| fs|, supt∈Fk

| ft |)∥∥∥

Lp≤ a .

This implies ‖max(| fs|,g)‖Lp ≤ a = ‖g‖Lp , so that | fs| ≤ g a.e. for all s ∈ R.]

4.3.8. (E. Prestini) Show that for f ∈ L2(T2) we have that

∑|m1|≤N

|m2|≤N2

f (m1,m2)e2πi(m1x1+m2x2)→ f (x1,x2)

for almost all (x1,x2) in T2.[Hint: Use the splitting f (m1,m2) = f (m1,m2)χ|m2|≤|m1|2 + f (m1,m2)χ|m2|>|m1|2

and apply the idea of the proof of Theorem 4.3.16.]

4.4 Applications to Geometry and Partial Differential Equations

In this section we discuss two applications of Fourier series. The first concerns a

classical result in planar geometry and the other the heat equation.

4.4.1 The Isoperimetric Inequality

Suppose we are given a closed positively oriented nonself intersecting C 1 curve C

in the (x,y) plane of length L that encloses a region R of area A. The curve can be

described in terms of its parametric equations x = x(t) and y = y(t), where t ∈ [0,1].Since the curve is closed, we have (x(0),y(0)) = (x(1),y(1)) and the C 1 functions

x(t),y(t) can be thought of as 1-periodic functions on the circle. The perimeter L of

the curve is given by the equation

L =

∫ 1

0

√|x′(t)|2 + |y′(t)|2 dt

while the area of the region R enclosed by the curve is equal to

A =∫∫

R1dxdy

=∫∫

R

∂

∂x

x

2− ∂

∂y

(− y

2

)dxdy

=1

2

∮

Cxdy− ydx

=1

2

∫ 1

0x(t)y′(t)− x′(t)y(t)dt ,

where we made use of Green’s theorem in the third equality above.

4.4 Applications to Geometry and Partial Differential Equations 293

A C 1 curve γ(t) is regular if γ ′(t) = 0 for all t. We have the following result

relating the perimeter and the enclosed area of a region enclosed by a closed C 1

curve.

Theorem 4.4.1. Given a closed, positively oriented, nonself intersecting, regular,

C 1 planar curve of length L that encloses a region of area A, we have that

A≤ L2/4π (4.4.1)

with equality holding if and only if the curve is a circle.

Proof. Assume that the curve has parametric equations x = x(t), y = y(t), 0≤ t ≤ 1.

We may assume that the curve has constant speed, i.e., it satisfies

√|x′(t)|2 + |y′(t)|2 = L

for all t ∈ [0,1]. This is achieved via the reparametrization of the curve in terms of

the inverse function s−1(t) = γ(t) of the normalized arc length function

s(t) =1

L

∫ t

0

√|x′(u)|2 + |y′(u)|2 du .

Since |(x′(t),y′(t))| = 0, t → s(t) is a one-to-one and onto continuous map from

[0,1] to [0,1]. Then the curve t → (x(γ(t)),y(γ(t))) has constant speed, since

|x′(γ(t))|2|γ ′(t)|2 + |y′(γ(t))|2|γ ′(t)|2 = |x′(s−1(t))|2 + |y′(s−1(t))|2|s′(s−1(t))|2 = L2 .

So we can replace the map (x(t),y(t)) by (x(γ(t)),y(γ(t))) which produces the same

curve. Let

f (t) = x(t)+ iy(t)

for t ∈ [0,1]. Then in view of the preceding discussion, we may assume that the

function f (t) = x(t)+ iy(t) satisfies | f ′(t)|= L for all t ∈ [0,1].Under the assumption | f ′(t)|= L for all t ∈ [0,1], we now show that (4.4.1) holds,

with equality if and only if f (t) = c0e2πit +C0 for some c0,C0 ∈ C with |c0| = L2π .

To prove this claim we argue as follows:

A =1

2Im

∫ 1

0f ′(t) f (t)dt

=1

2Im

∫ 1

0f ′(t)

(f (t)− f (0)

)dt

≤ 1

2L∥∥ f − f (0)

∥∥L2

≤ L

2

1

2π

∥∥ f ′∥∥

L2

=L2

4π,


establishing (4.4.1), but we need to explain why the inequality

‖ f − f (0)‖L2 ≤ 1

2π‖ f ′‖L2 (4.4.2)

is valid. Indeed, we have

f ′(t) = ∑m∈Z

2πim f (m)e2πimt

where the series converges in L2. Thus we have

∥∥ f ′∥∥

L2 = 2π

[∑

m∈Z

|m f (m)|2] 1

2

≥ 2π

[∑

m∈Z\0| f (m)|2

] 12

= 2π∥∥ f− f (0)

∥∥L2 , (4.4.3)

which proves (4.4.2).

Now suppose that equality holds in (4.4.1), then we must have equality in (4.4.2)

and thus in (4.4.3), which implies that f (m) = 0 when |m| ≥ 2; hence for all t ∈ [0,1]we must have

f (t) = ce2πit + c′e−2πit + f (0) (4.4.4)

where c,c′ are complex numbers. But since ‖ f ′‖L2 = L, it follows that

4π2(|c|2 + |c′|2) = L2 , (4.4.5)

and since | f ′(t)|= L for all t ∈ [0,1], it follows that

( L

2π

)2

= |c|2 + |c′|2−2Re[cc′e2πi2t

](4.4.6)

for all t ∈ [0,1]. Combining (4.4.5) and (4.4.6) we obtain

Re[cc′e2πi2t

]= 0 . (4.4.7)

Inserting t = 0 and t = 1/8 in (4.4.7) and using that Im(iz) =−Rez, we deduce that

Recc′ = Imcc′ = 0. This implies that either c or c′ is zero. In either case (4.4.4) and

(4.4.5) imply that f (t) is a circle of radius L/2π centered at the point f (0).

4.4.2 The Heat Equation with Periodic Boundary Condition

Let k > 0 be a fixed quantity. Consider the partial differential equation

∂

∂ tF(x, t) = k

n

∑j=1

∂ 2

∂x2j

F(x, t) t ∈ (0,∞), x ∈ Rn , (4.4.8)


which is called the heat equation. Assume that there is an initial condition

F(0,x) = f (x) x ∈ Rn (4.4.9)

for a given C ∞ function f on Rn which is assumed to be 1-periodic in every variable.

We would like to find a continuous function F(t,x) on [0,∞)×Rn which is C ∞

on (0,∞)×Rn such that

F(t,x+ e j) = F(t,x)

for all t ≥ 0 and all e j = (0, . . . ,0,1,0, . . . ,0), so that F solves the equation (4.4.8).

The function F(t,x) represents the temperature of a body at time t > 0 at the

location (x1, . . . ,xn). Since the initial temperature f is 1-periodic in each variable,

we expect F(t, ·) to also be periodic in each variable. For example, F(t,x) is a good

model for the temperature of the torus (e2πix1 , . . . ,e2πixn) : x j ∈ R at time t > 0,

given that its temperature at time t = 0 is f (x). When n = 1, F(t,x) models the

temperature of a infinitesimally thin ring, thought of as the unit circle, at time t > 0

at the location e2πix.

Let us suppose there is a continuous function F(t,x) on [0,∞)×Rn which is C ∞

on (0,∞)×Rn that solves the equation (4.4.8) and satisfies F(t,x+ e j) = F(t,x)for all x ∈ Rn and t ≥ 0. Denote by cm(t) the Fourier coefficient of the function

x → F(t,x) defined by

cm(t) =∫

TnF(t,x)e−2πim·x dx .

Then cm(t) is a continuous function on [0,∞) since F is continuous in the variable

t. For the same reason, cm is a smooth function on (0,∞) whose jth derivative is

given by

d j

dt jcm(t) =

∫

Tn

∂ j

∂ t jF(t,x)e−2πim·x dx

for any j = 1,2, . . . . Using equation (4.4.8) we obtain that

c′m(t) =∫

Tn

∂

∂ tF(t,x)e−2πim·x dx =

∫

Tnk∂ 2

∂ 2xF(t,x)e−2πim·x dx =−4π2|m|2k cm(t),

where the last identity is due to an integration by parts in which the boundary terms

cancel each other in view of the periodicity of the integrand in x. Also cm(0) = f (m).The ordinary differential equation c′m(t) = −4π2|m|2k cm(t) with initial condition

cm(0) = f (m) is easily solved by separating the variables

dcm(t)

cm(t)=−4π2|m|2kdt , (4.4.10)

yielding the solution

cm(t) = f (m)e−4π2|m|2kt .


We may therefore define the function

F(t,x) = ∑m∈Zn

f (m)e−4π2|m|2kte2πim·x (4.4.11)

on [0,∞)×Rn and observe the following:

(a) F is continuous on [0,∞)×Rn and C ∞ on (0,∞)×Rn.

(b) F satisfies the heat equation (4.4.8) and the initial condition (4.4.9).

(c) F is 1-periodic in each of the last n variables.

These statements can be easily proved by passing the differentiation inside the

sum, in view of the rapid convergence of the series in (4.4.11) due to the fact that

the periodic function f is C ∞(Rn). Furthermore, F is unique with properties (a),

(b), and (c), since any other function G(x, t) with these properties is derived in the

preceding way, and so it has to be equal to F(x, t).

Definition 4.4.2. Define the heat kernel

Ht(x) = ∑m∈Zn

e−4π2|m|2kte2πim·x

for t > 0. Notice that the series defining Ht is absolutely convergent for any t > 0.

The importance of the heat kernel lies in the fact that one can express the solution

F(x, t) of (4.4.8) in terms of the convolution F(x, t) = ( f ∗Ht)(x).

We summarize these facts in the following proposition.

Proposition 4.4.3. Let k > 0 be fixed and let f be in C ∞(Rn). Assume that f is

1-periodic function in each variable. Then the heat equation

∂

∂ tF(x, t) = k∆xF(x, t) t ∈ (0,∞), x ∈ Rn (4.4.12)

under the initial condition

F(0,x) = f (x) x ∈ Rn (4.4.13)

has a unique solution which is continuous on [0,∞)×Rn and C ∞ on (0,∞)×Rn

given by

F(x, t) = ( f ∗Ht)(x) = ∑m∈Zn

f (m)e−4π2|m|2kte2πim·x . (4.4.14)

Proof. Since f is C ∞, the series in (4.4.14) is rapidly convergent in m and thus it

gives a continuous function on [0,∞)×Rn. Moreover, the series can be differentiated

term by term in the variable t > 0, and thus it produces a C ∞ function on (0,∞)×Rn.

By Fourier inversion (Proposition 3.2.5), F satisfies the initial condition (4.4.13).

Finally, to verify (4.4.12), we simply notice that

∂

∂ tF(x, t) = ∑

m∈Zn

∂

∂ tf (m)e−4π2|m|2kte2πim·x

= −4π2k ∑m∈Zn

f (m)e−4π2|m|2kt |m|2e2πim·x

= k ∑m∈Zn

f (m)e−4π2|m|2kt( ∂ 2

∂ 2x1

+ · · ·+ ∂ 2

∂ 2xn

)e2πim·x

= k( ∂ 2

∂ 2x1

+ · · ·+ ∂ 2

∂ 2xn

)∑

m∈Zn

f (m)e−4π2|m|2kte2πim·x

= k∂ 2

∂ 2xF(x, t) ,

where the rapid convergence of the series in m makes it possible to pass the differ-

entiations in and out of the sum. Finally, to show uniqueness, assume that there is

another solution G(t,x), continuous on [0,∞)×Rn and C ∞ on (0,∞)×Rn, that can

be expanded in Fourier series as follows:

G(t,x) = ∑m∈Zn

cm(t)e2πim·x .

Conditions (4.4.12) and the rapid decay of the coefficients cm(t) yield the ordinary

differential equation (4.4.10) with initial condition cm(0) = f (m), which has the

solution cm(t) = f (m)e−4π2|m|2kt . Thus G = F on [0,∞)×Rn.

It is important to observe that the family Htt>0 is an approximate identity on

T1. Indeed, the Poisson summation formula (Theorem 3.2.8) and the fact that the

inverse Fourier transform of e−4π2kt|ξ |2 is e−|x|2/4kt/(2

√πkt)n [Example 2.2.9 and

Proposition 2.2.11 (8)] yield that for all x ∈ [0,1]n we have

Ht(x) =1

(2√πkt)n ∑

ℓ∈Z

e−|x+ℓ|2

4kt .

This identity implies that Ht(x)≥ 0 for all t > 0 and that

∫

TnHt(x)dx =

∫

Rn

1

(2√πkt)n

e−|x|24kt dx =

∫

Rne−π|x|

2

dx = 1

for all t > 0 and that

∫

δ≤|u|Ht(u)du≤

∫

|x|≥δ1

(2√πkt)n

e−|x|24kt dx =

∫

|x|≥ δ2√

kπt

e−π|x|2

dx

which tends to zero as t → 0 for any δ > 0 in view of the Lebesgue differentiation

theorem. Thus properties (i), (ii), and (iii) of approximate identities hold.

As a consequence, we have that ‖F(t, ·)− f‖Lp(Tn) → 0 as t → 0 for 1 ≤ p < ∞and F(t, ·) converges to f uniformly on Tn; see Theorem 1.2.19.


Exercises

4.4.1. Let f ,F be as in Proposition 4.4.3. Prove that the total heat on the torus

remains constant in time by showing that for all t ≥ 0 we have

∫

TnF(t,x)dx =

∫

Tnf (x)dx .

Moreover, show that the temperature at any fixed point x ∈ Tn on the torus tends to

the average initial temperature, i.e., it satisfies

limt→∞

F(t,x) =∫

Tnf (y)dy .

4.4.2. Derive the following property of the heat kernel,

Ht ∗Hs = Ht+s

for all t,s > 0.

4.4.3. Consider the heat equation on [0,∞)×R

∂

∂ tu(x, t) =

∂ 2

∂x2u(x, t)

without a boundary condition. Show that u = 2t + x2 and u(t,x) = e−q2teiqx, as well

as constant functions, are solutions of this equation. Prove that the set of solutions

is a vector space over the field of complex numbers C.

4.4.4. Suppose that a square-integrable function g(x) on Rn is supported in a cube

[−A,A]n for some A > 0. Then we have the following representation:

g(x) = ∑m∈Zn

(1

(2A)n

∫

Rng(y)e−2πi

m·y2A dy

)e2πi x·m

2A χ[−A,A]n ,

where the series converges in L2

4.4.5. This exercise provides an application of Fourier series in complex analysis.

Let z ∈C\Z. Consider the function hz(x) = cos(2πzx) defined on [− 12, 1

2] extended

periodically on the entire line [notice hz(− 12) = hz(

12)].

(a) Compute the Fourier coefficients of hz.

(b) Obtain a Fourier series expansion of hz noticing that it is a Lipschitz function.

(c) Plug in x = 1/2 to prove that

cot(πz) =1

πz+

1

π

∞

∑m=1

2z

z2−m2.

4.5 Applications to Number theory and Ergodic theory 299

4.5 Applications to Number theory and Ergodic theory

In this section we discuss three applications of Fourier series techniques to number

theory and ergodic theory.

4.5.1 Evaluation of the Riemann Zeta Function at even Natural

numbers

Definition 4.5.1. We define the Bernoulli polynomials Bk∞k=0 on [0,1] recursively

as follows:

B0(x) = 1

B′k(x) = kBk−1(x)

for k = 1,2, . . . , and ∫ 1

0Bk(x)dx = 0 .

In view of this definition we find the first few polynomials B1(x) = x− 12, B2(x) =

x2−x+ 16, B3(x) = x3− 3

2x2+ 1

2x, etc. Unlike orthogonal polynomials, the Bernoulli

polynomials have the remarkable property that their number of zeros in the unit

interval does not increase as the degree of the polynomials increases; in fact all

Bernoulli polynomials have at most three zeros in [0,1].Notice that for k ≥ 2 we have

Bk(1)−Bk(0) =∫ 1

0B′k(x)dx = k

∫ 1

0Bk−1(x)dx = 0 ,

thus we may think of these polynomials as functions on the circle T1. We extend the

Bernoulli polynomials to the whole line periodically by setting Bk(x+ l) = Bk(x) for

x ∈ [0,1]. We now compute the Fourier coefficients of Bk. We have

B1(m) =∫ 1

0(t− 1

2)e−2πimt dt =

0 if m = 0

− 12πim

if m = 0.

Therefore, using Corollary 3.4.10, we can write

B1(x) = ∑m=0

− 1

2πime2πimx

where the series converges at every x ∈ (0,1).We have the following result concerning the Fourier expansion of the Bernoulli

polynomials.


Theorem 4.5.2. For each k ≥ 2 we have

Bk(x) =−k! ∑m∈Z\0

1

(2πim)ke2πimx , (4.5.1)

where the series converges absolutely and uniformly on [0,1]. When k = 1 we have

B1(x) =− ∑m∈Z\0

1

2πime2πimx (4.5.2)

for all x ∈ (0,1) and the series converges conditionally.

Proof. We have already proved (4.5.2) and we focus attention to the case k ≥ 2. As

a consequence of B′k = kBk−1 we obtain

Bk(x) = k

∫ x

0Bk−1(t)dt +Ck .

Using the property that Bk has integral zero over [0,1] we evaluate the constant Ck.

We have

0 =∫ 1

0

[k

∫ x

0Bk−1(t)dt +Ck

]dx

= k

∫ 1

0

(∫ 1

tdx

)Bk−1(t)dt +Ck

= −k

∫ 1

0tBk−1(t)dt +Ck .

Thus

Ck = k

∫ 1

0tBk−1(t)dt .

The Fourier series of Bk(x) can be obtained by integrating the one for Bk−1(x)for all k ≥ 2 by induction via the identity

Bk(x) = k

∫ x

0Bk−1(t)dt + k

∫ 1

0tBk−1(t)dt . (4.5.3)

Indeed, assume that (4.5.1) holds for some k ≥ 2. Then using (4.5.3) we obtain

Bk(x) =∫ x

0lim

N→∞∑|m|≤Nm=0

−k(k−1)!

(2πim)k−1e2πimt dt−

∫ 1

0t lim

N→∞∑|m|≤Nm=0

k(k−1)!

(2πim)k−1e2πimt dt

= limN→∞

∫ x

0∑|m|≤Nm=0

−k(k−1)!

(2πim)k−1e2πimt dt− lim

N→∞

∫ 1

0t ∑|m|≤Nm=0

k(k−1)!

(2πim)k−1e2πimt dt


= − limN→∞

∑|m|≤Nm=0

k!

(2πim)k−1

e2πimx−1

2πim− lim

N→∞∑|m|≤Nm=0

k!

(2πim)k−1

[te2πimt

2πim

]t=1

t=0

= − k! ∑m=0

e2πimx

(2πim)k.

Passing the limit from inside the integral to outside is allowed due to the uniform

convergence of the series when k≥ 3. In the case k = 2, one may use Exercise 3.5.6

which says that for all [a,b] T1 and g integrable functions over [a,b] one has

limN→∞

∫ b

a(g∗DN)(t)dt =

∫ b

alim

N→∞(g∗DN)(t)dt =

∫ b

ag(t)dt .

This argument proves identity (4.5.1) for all k ≥ 2 by induction and concludes the

proof.

We recall the following definition from number theory.

Definition 4.5.3. For s > 1 we define

ζ (s) =∞

∑k=1

1

ks

called the Riemann zeta function.

We use the Fourier expansions of the Bernoulli polynomials to obtain the values

of the Riemann zeta function for integers. When k is an even integer, identity (4.5.1)

can also be written as

Bk(x) = 2(−1)1+ k2 k!

∞

∑n=1

cos(2πnx)

(2πn)k

and inserting x = 0 yields

ζ (k) =∞

∑n=1

1

nk=

Bk(0)(2π)k

2(−1)1+ k2 k!

.

The polynomial B1(x) = x− 1/2 has rational coefficients and thus so do all the Bk

by a straightforward inductive argument that uses the identity (4.5.3). Thus Bk(0) is

a rational number for all k ≥ 1. We conclude that

ζ (2m) =B2m(0)(2π)

2m

2(−1)1+m(2m)!(4.5.4)

which is a rational multiple of (2π)2m, hence transcendental, since π is a transcen-

dental number. We have therefore obtained the following.

Corollary 4.5.4. (Euler) The value of the Riemann zeta function ζ (2m), m =1,2, . . . , is equal to a rational multiple of (2π)2m; hence it is a transcendental

number.

The corresponding statement for odd integers remains unresolved in general, as

of this writing.

4.5.2 Equidistributed sequences

Here we discuss Weyl’s theorem on equidistributed sequences.

Definition 4.5.5. A sequence ak∞k=0 with values in Tn is called equidistributed if

for every cube Q in Tn we have

limN→∞

#k : 0≤ k ≤ N−1, ak ∈ QN

= |Q| .

Theorem 4.5.6. The following statements are equivalent:

(a) The sequence ak∞k=0 is equidistributed.

(b) For every smooth function f on Tn we have that

limN→∞

1

N

N−1

∑k=0

f (ak) =∫

Tnf (x)dx .

(c) For every m ∈ Zn \0 we have

limN→∞

1

N

N−1

∑k=0

e2πim·ak = 0 .

Proof. We first prove the equivalence of (a) and (b). We begin by observing that

(b) is a restatement of (a) if f = χQ and Q is a cube in Tn. Thus, if (a) holds,

then (b) holds for all step functions, i.e., finite linear combinations of characteristic

functions of cubes. We prove that (a) implies (b) for smooth functions. Given a

smooth function f on Tn and given ε > 0, by the uniform continuity of f , there is a

step function g =∑mj=1 c jχQ j

(c j ∈C and Q j are cubes in Tn) such that ‖ f −g‖L∞ <ε3. Since g is a finite linear combination of step functions, there is an N0 such that

for N ≥ N0 we have ∣∣∣∣1

N

N−1

∑k=0

g(ak)−∫

Tng(x)dx

∣∣∣∣<ε

3.

Since ∣∣∣∣∫

Tnf (x)dx−

∫

Tng(x)dx

∣∣∣∣≤ ‖ f −g‖L∞ <ε

3

and ∣∣∣∣1

N

N−1

∑k=0

g(ak)−1

N

N−1

∑k=0

f (ak)

∣∣∣∣≤ ‖ f −g‖L∞ <ε

3,

it follows that for N ≥ N0 we have

∣∣∣∣1

N

N−1

∑k=0

f (ak)−∫

Tnf (x)dx

∣∣∣∣< ε ,

thus (b) holds.

To prove that (b) implies (a) given a cube Q in Tn pick two smooth functions g

and h such that

0≤ h≤ χQ ≤ g

and such that g is equal to 1 on Q and vanishes off (1+ε)Q while h is equal to 1 on

(1− ε)Q and vanishes off Q. Observe that

|Q|− cn ε ≤∫

Tnh(x)dx≤ |Q| ≤

∫

Tng(x)dx≤ |Q|+ cn ε .

for some cn > 0. Since

1

N

N−1

∑k=0

h(ak)≤1

N

N−1

∑k=0

χQ(ak)≤1

N

N−1

∑k=0

g(ak) ,

the sandwich theorem implies that

|Q|− cn ε ≤ liminfN→∞

1

N

N−1

∑k=0

χQ(ak)≤ limsupN→∞

1

N

N−1

∑k=0

χQ(ak)≤ |Q|+ cn ε .

Since ε > 0 was arbitrary the conclusion follows.

The implication (b) =⇒ (c) is trivial.

We now prove that (c) =⇒ (b).Given a smooth function f on Tn we write

1

N

N−1

∑k=0

f (ak) =1

N

N−1

∑k=0

∑m∈Zn

f (m)e2πim·ak = f (0)+ ∑m∈Zn\0

f (m)

(1

N

N−1

∑k=0

e2πim·ak

).

Because of the rapid decay of the Fourier coefficients of f we can pass the limit as

N → ∞ inside the sum in m. It follows that

limN→∞

1

N

N−1

∑k=0

f (ak) = f (0) =∫

Tnf (x)dx .

Example 4.5.7. The sequence k√

2− [k√

2]∞k=0 is equidistributed on T1. We check

this by verifying condition (c) of Theorem 4.5.6. Indeed if m ∈ Z\0 then

limN→∞

1

N

N−1

∑k=0

e2πim(k√

2−[k√

2]) = limN→∞

1

N

e2πiN(m√

2)−1

e2πi(m√

2)−1= 0 ,

since m√

2 is never a rational and thus the denominator never vanishes.

Naturally, the same conclusion is valid for any other irrational number in place

of√

2.

Example 4.5.8. We examine the sequence of the first digits of powers of 2. Consider

the following sequence of numbers defined for m = 1,2, . . .

dm = first digit of 2m.

For instance we have d1 = 2,d2 = 4,d3 = 8,d4 = 1,d5 = 3, . . . .Fix an integer k ∈ 1,2,3,4,5,6,7,8,9. We would like to find the frequency in

which k appears as a first digit of 2m, precisely, we would like to compute

limN→∞

#

m ∈ 1,2, . . . ,N : dm = k

N.

The crucial observation is that the first digit of 2m is equal to k if and only if there is

a nonnegative integer s such that

k10s ≤ 2m < (k+1)10s .

Taking logarithms with base 10 we obtain

s+ log10(k)≤ m log10 2 < s+ log10(k+1) ,

but since 0≤ log10(k) and log10(k+1)≤ 1, taking fractional parts we obtain that

s = [m log10 2]

and that

log10(k)≤ m log10 2− [m log10 2]< log10(k+1) .

Since the number log10 2 is irrational, it follows from Example 4.5.7 that the

sequence m log10 2− [m log10 2]

∞m=1

is equidistributed in [0,1). Using Definition 4.5.5 in dimension n = 1 with

Q = [a,b] =[

log10(k), log10(k+1)]

we obtain that

limN→∞

#

m ∈ 1,2, . . . ,N : dm = k

N= log10(k+1)− log10(k) = log10(1+

1k) .

This gives the frequency in which k appears as first digit of 2m. Notice that

9

∑k=1

log10(1+1k) =

9

∑k=1

(log10(k+1)− log10(k)

)= 1 ,

as expected, and that the digit with the highest frequency that appears first in a

term of the sequence 1,2,4,8,16,32,64, . . . is 1, while the one with the lowest

frequency is 9.

4.5.3 The Number of Lattice Points inside a Ball

Points in Zn are called lattice points. In this subsection we obtain the number of

lattice points N(R) inside a closed ball of radius R in Rn centered at the origin,

precisely, we compute the asymptotic behavior of

N(R) = |B(0,R)∩Zn|

as R→ ∞. We denote by vn the volume of the closed unit ball in Rn. We have the

following result.

Theorem 4.5.9. Let n ≥ 2. If N(R) is the number of lattice points inside the closed

ball of radius R centered at zero in Rn, then we have that

N(R) = vnRn +O(Rn n−1n+1 ) ,

as R→ ∞.

Proof. Let B be the closed unit ball in Rn and χB its characteristic function. Using

the result in Appendix B.5 we have χB(ξ ) =Jn/2(2π|ξ |)|ξ |n/2 . Now in view of the behavior

of the Bessel function given in Appendix B.6 for |ξ | < 12π we have Jn/2(2π|ξ |) ≤

C|ξ | n2 . Also for |ξ | ≥ 1

2π we have Jn/2(2π|ξ |) ≤ C|ξ |− 12 , in view of the result in

Appendix B.7. Consequently, there is a constant Cn such that for all ξ ∈Rn we have

|χB(ξ )| ≤Cn(1+ |ξ |)−n+1

2 .

Fix a smooth nonnegative radial function ζ supported in |x| ≤ 12

with integral

equal to 1 and define ζε(x) =1εn ζ (

xε ) for ε > 0. For 0 < ε < 1

10, define functions

Φε = χ(1− ε

2)B∗ζε

Ψ ε = χ(1+

ε2)B∗ζε .

These functions are even, hence their Fourier transforms are real-valued. We observe

that

Φε(x) = 1 when |x| ≤ 1− ε and Φε(x) = 0 when |x| ≥ 1. (4.5.5)

Indeed, we have

Φε(x) =

∫

Rnχ(1− ε

2 )B(y)ζε(x− y)dy

=∫

|y|≤1− ε2

1

εnζ ( x−y

ε )dy

=∫

|y|≤ 1ε− 1

2

ζ ( xε − y)dy .

For |x| ≤ 1−ε , | xε | ≤ 1ε −1, so | xε − t| ≤ 1

ε −1+ 12= 1

ε − 12

for |t| ≤ 12, which means

Φε(x) =∫

|y|≤ 1ε− 1

2

ζ ( xε − y)dy =

∫

|t|≤ 12

ζ (t)dt = 1.

For |x| ≥ 1, | xε − y| ≥ 1ε − 1

ε +12= 1

2, so

Φε(x) =∫

|y|≤ 1ε− 1

2

ζ ( xε − y)dy = 0,

proving (4.5.5). Likewise one can show that

Ψ ε(x) = 1 when |x| ≤ 1 and Ψ ε(x) = 0 when |x| ≥ 1+ ε . (4.5.6)

Next we claim that

∣∣Φε(ξ )∣∣+

∣∣Ψ ε(ξ )∣∣≤Cn,N(1+ |ξ |)−

n+12 (1+ ε |ξ |)−N (4.5.7)

for every ξ ∈ Rn and N a large positive number. Indeed to show (4.5.7) for Φε we

write

|Φε(ξ )| = |χ(1− ε2 )B

(ξ )ζ (ξε)|

≤ (1− ε2)nCn(1+ |ξ |(1− ε

2))−

n+12 |ζ (ξε)|

≤Cn,N(1+ |ξ |)−n+1

2 (1+ ε |ξ |)−N

(4.5.8)

since ζ ∈S (R) and 0 < ε < 110

. The proof forΨ ε is completely similar.

We now notice that for R > 0, m ∈ Zn \0, and x ∈ [0,1]n we have

1+ |m+ x|R≤ 1+(√

n+ |m|)R≤ 2√

n(1+ |m|R) .

This implies that for ε < 1/10 we have

∑m∈Zn\0

Rn(1+R|m|)− n+12 (1+ εR|m|)−N

≤C′∫

[0,1]n∑

m∈Zn\0Rn(1+R|m+ x|)− n+1

2 (1+ εR|m+ x|)−Ndx

≤C′∫

RnRn(1+R|x|)− n+1

2 (1+ εR|x|)−Ndx

≤C′′ε−n−1

2 , (4.5.9)

where the proof of (4.5.9) is easily deduced by considering the cases (a) |x| ≤ R−1

which yields a constant, (b) R−1 ≤ |x| ≤ (Rε)−1 which yields a constant multiple of

ε−n−1

2 , and (c) (Rε)−1 ≤ |x| which also produces a constant multiple of ε−n−1

2 if we

pick N > n−12

.

Using (4.5.5) and the Poisson summation formula we write

∑m∈Zn

χB(mR)≥ ∑

m∈Zn

Φε(mR)

= RnΦε(0)+ ∑m∈Zn\0

RnΦε(Rm)

≥ vnRn(1− ε)n−Cn,N ∑m∈Zn\0

Rn(1+R|m|)− n+12 (1+ εR|m|)−N

≥ vnRn−nvnRnε−C′n,Nε− n−1

2 ,

where we used that (1− ε)n ≥ 1− nε , (4.5.8), and (4.5.9). Now pick ε such that

εRn = ε−n−1

2 , or equivalently ε = R−2n

n+1 to deduce the estimate

N(R)≥ vnRn−O(Rn n−1n+1 )

as R→ ∞.

Finally, making use of (4.5.6), and via a similar argument we write

∑m∈Zn

χB(mR)≤ ∑

m∈Zn

Ψ ε(mR) = RnΨ ε(0)+ ∑

m∈Zn\0RnΨ ε(Rm)

≤ vnRn(1+ ε)n +Cn,N ∑m∈Zn\0

Rn(1+R|m|)− n+12 (1+ εR|m|)−N

≤ vnRn + vn 2n Rnε+Cn,Nε− n−1

2 .

The same choice of ε = R−2n

n+1 , yields the upper estimate for N(R).

Combining the upper and lower estimates for N(R) we obtain

N(R) = vnRn +O(Rn n−1n+1 ) ,

as R→ ∞.

Exercise 4.5.8 contains an application of Theorem 4.5.9.

Exercises

4.5.1. Prove that for all x ∈ [0,1] we have

∞

∑j=1

sin(2π jx)

j2m+1=

(−1)m+1

2

(2π)2m+1

(2m+1)!B2m+1(x) .

4.5.2. Show that for all z ∈ C with |z|< 1 we have

πzcot(πz) = 1−2∞

∑k=0

z2k+2ζ (2k+2) .

[Hint: Use the result of Exercise 4.4.5.

]

4.5.3. Suppose that a point x=(x1, . . . ,xn)∈ [0,1]n has the property that m ·x is irra-

tional for all m∈Zn\0. Show that the sequence (kx1− [kx1], . . . ,kxn− [kxn])∞k=0

is equidistributed in Tn.

4.5.4. ([191]) Let N(x,R) be the number of lattice points inside the closed ball of

radius R > 0 centered at x ∈ Rn. Show that

∫

Tn

∣∣N(x,R)− vnRn∣∣2 dx = O(Rn−1)

as R→ ∞, where vn is the volume of the unit ball on Rn.

4.5.5. (Minkowski) Let S be an open convex symmetric set in Rn and assume that

the Fourier transform of its characteristic function satisfies the decay estimate

|χS(ξ )| ≤C(1+ |ξ |)− n+12 .

(This is the case if the boundary of S has nonzero Gaussian curvature.) Assume that

|S|> 2n. Prove that S contains at least one lattice point other than the origin.[Hint: Assume the contrary, set f = χ 1

2 S∗ χ 1

2 S, and apply the Poisson summation

formula to f to prove that f (0)≥ f (0).]

4.5.6. For t ∈ [0,∞) let

N(t) = #m ∈ Zn : |m| ≤ t .

Let 0 = r0 < r1 < r2 < · · · be the sequence all of numbers r for which there exist

m ∈ Zn such that |m|= r.

(a) Observe that N is right continuous and constant on intervals of the form [r j,r j+1).(b) Show that the distributional derivative of N is the measure

µ(t) = #m ∈ Zn : |m|= t ,

defined via the identity 〈µ ,ϕ〉= ∑∞j=0 #m ∈ Zn : |m|= r jϕ(r j).

4.5.7. Let f ∈ C 1((0,∞)), and let 0 < a < b < ∞. Derive the identity

∑m∈Zn

a<|m|≤b

f (|m|) =∫ b

af (t)dN(t) = f (b)N(b)− f (a)N(a)−

∫ b

af ′(x)N(x)dx ,

where N is defined in Exercise 4.5.6 and∫ b

a f (t)dN(t) is the Riemann-Stieltjes in-

tegral of f with respect to N.

4.5.8. Let n ∈ Z+ and 0 < λ < ∞.

(a) Prove that for k ∈ Z+∪0 we have

∑m∈Zn

a<|m|≤b

ei|m|

|m|λ =−iωn−1 eib

bλ−(n−1)− −iωn−1 eia

aλ−(n−1)+O(a−λ+(n−1)− n−1

n+1 )

for all 0 < a n− n−1n+1

, the limit

limR→∞

∑m∈Zn\0|m|≤R

ei|m|

|m|λ

exists.

(c) Prove, however, that when n−1− n−1n+1

< λ ≤ n−1, the limit in part (b) does not

exist.[Hint: Use Exercise 4.5.7 and Theorem 4.5.9. Part (b): For R > 1 use the identity

∑m∈Zn\0|m|≤R

ei|m|

|m|λ =[R− 1

2 ]−1

∑k=0

(∑

m∈Zn

k+ 12<|m|≤k+ 3

2

ei|m|

|m|λ

)+

∑

m∈Zn

[R− 12 ]+

12<|m|≤R

ei|m|

|m|λ

.

Notice that the main term in the first sum on the right is telescoping. Part (c): Show

that limR→∞∑ m∈Zn\0R<|m|≤R+1

ei|m|

|m|λ does not tend to zero.]

HISTORICAL NOTES

The boundedness of the conjugate function on the circle (Theorem 4.1.7) and, hence, the Lp

convergence of one-dimensional Fourier series was announced by Riesz in [292], but its proof ap-

peared a little later in [293]. The proof of Theorem 4.1.7 in the text is attributed to S. Bochner.

Luzin’s conjecture [235] on almost everywhere convergence of the Fourier series of continuous

functions was announced in 1913 and settled by Carleson [54] in 1965 for the more general class

of square summable functions (Theorem 4.3.14). Carleson’s theorem was later extended by Hunt

[165] for the class of Lp functions for all 1 < p < ∞ (Theorem 4.3.15). Sjolin [325] sharpened

this result by showing that the Fourier series of functions f with | f |(log+ | f |)(log+ log+ | f |) in-

tegrable over T1 converge almost everywhere. Antonov [5] improved Sjolin’s result by extend-

ing it to functions f with | f |(log+ | f |)(log+ log+ log+ | f |) integrable over T1. One should also

consult the related results of Soria [330] and Arias de Reyna [9]. The book [10] of Arias de

Reyna contains a historically motivated comprehensive study of topics related to the Carleson–

Hunt theorem. Counterexamples due to Konyagin [200] show that Fourier series of functions f

with | f |(log+ | f |) 12 (log+ log+ | f |)− 1

2−ε integrable over T1 may diverge when ε > 0. Examples of

continuous functions whose Fourier series diverge exactly on given sets of measure zero are given

in Katznelson [189] and Kahane and Katznelson [183].

The extension of the Carleson–Hunt theorem to higher dimensions for square summability of

Fourier series (Theorem 4.3.16) is a rather straightforward consequence of the one-dimensional

result and was independently obtained by Fefferman [112], Sjolin [325], and Tevzadze [359]. An

example showing that the circular partial sums of a Fourier series may not converge in Lp(Tn) for

n ≥ 2 and p = 2 was obtained by Fefferman [113]. This example also shows that there exist Lp

functions on Tn for n ≥ 2 whose circular partial sums do not converge almost everywhere when

1≤ p < 2. Indeed, if the opposite happened, then the maximal operator f → supN≥0 |D(n,N)∗ f |would have to be finite a.e. for all f ∈ Lp(Tn), and by Stein’s theorem [335] it would have to be of

weak type (p, p) for some 1 < p < 2. But this would contradict Fefferman’s counterexample on Lp1

for some p < p1 < 2. On the other hand, almost everywhere is valid for the square partial sums of

functions f with | f |(log+ | f |)n(log+ log+ log+ | f |) integrable over Tn, as shown by Antonov [6];

see also Sjolin and Soria [327].

The development of the complex methods in the study of Fourier series was pioneered by

the Russian school, especially Luzin and his students Kolmogorov, Menshov, and Privalov. The

existence of an integrable function on T1 whose Fourier series diverges almost everywhere (The-

orem 4.2.1) is due to Kolmogorov [195]. An example of an integrable function whose Fourier

series diverges everywhere was also produced by Kolmogorov [198] three years later. Localiza-

tion of the Bochner–Riesz means at the critical exponent α = n−12

fails for L1 functions on Tn

(see Bochner [30]) but holds for functions f such that | f | log+ | f | is integrable over Tn (see Stein

[333]). The latter article also contains the Lp boundedness of the maximal Bochner–Riesz oper-

ator supR>0 |BαR ( f )| for 1< p<∞ when α > | 1p− 1

2|. Proposition 4.1.9 is due to Stein [331] and

Theorem 4.2.5 is also due to Stein [335]. The technique that involves the points for which the set

|x−m| : m ∈ Zn is linearly independent over the rationals was introduced by Bochner [30].

Transference of regulated multipliers originated in the article of de Leeuw [94]. The methods

of transference in Section 4.3 were beautifully placed into the framework of a general theory by

Coifman and Weiss [70]. The key Lemma 4.3.8 is attributed to G. Weiss. Transference of maximal

multipliers (Theorem 4.3.12) was first obtained by Kenig and Tomas [192] and later elaborated by

Asmar, Berkson, and Gillespie [12], [13].

Paraphrasing Pappus of Alexandria, bees know than a hexagon will hold more honey than a

triangle or square of the same length, but people claim a greater share of wisdom knowing that the

circle of a given length holds the maximum area among all geometric shapes of equal perimeter.

This reflection captures the isoperimetric inequality, which was first recorded by Pappus in the

fourth century A.D. and was credited it to Zenodorus (second century B.C.). Archimedes also

studied the problem, but his work on the subject, like the original writings of Zenodorus, has been

lost. Rigorous modern-day proofs of this inequality can be traced to J. Steiner, K. Weierstrass, and


F. Edler, whose methods are based in geometry and calculus. The proof in the text is due to A.

Hurwitz. On the history of the isoperimetric inequality see [322].

The mean square error for lattice points (Exercise 4.5.4) is due to Kendall [191] while the more

delicate pointwise asymptotic formula of Theorem 4.5.9 was obtained by Landau [212]. Using

Landau’s formula Pinsky, Stanton, and Trapa [284] showed that the spherical partial sums of the

Fourier series of the characteristic function of a sufficiently small ball in Tn converge at the center

of the ball if and only if the dimension n is strictly less than three; this property is valid for the

characteristic function of any ball as shown in Pinsky [283].

Chapter 5

Singular Integrals of Convolution Type

The topic of singular integrals is motivated by its intimate connection with some of

the most important problems in Fourier analysis, such as that of the convergence of

Fourier series. As we have seen, the Lp boundedness of the conjugate function on

the circle is equivalent to the Lp convergence of Fourier series of Lp functions. And

since the Hilbert transform on the real line provides an analogue of the conjugate

function on the circle, it is deeply connected with the Lp convergence of Fourier

integrals. It also appears in the theory of harmonic functions on the upper half space

and has so many remarkable properties that deserve a careful investigation. The

Hilbert transform is the prototype of all singular integrals and provides inspiration

for subsequent development of the subject.

Historically, the theory of the Hilbert transform depended on techniques of com-

plex analysis. With the development of the Calderon–Zygmund school, and the

extension of one-dimensional theory to higher dimensions, real-variable methods

slowly replaced complex analysis. The higher-dimensional framework proved to be

flexible enough for generalizations and led to the introduction of singular integrals

in other areas of mathematics. Singular integrals are nowadays intimately connected

with partial differential equations, operator theory, several complex variables, and

other fields. In this chapter we study singular integrals given by convolution with

tempered distributions. We call such operators singular integrals of convolution

type.

5.1 The Hilbert Transform and the Riesz Transforms

We begin the investigation of singular integrals with a careful study of the Hilbert

transform which provides inspiration for the subsequent development of the theory.



313

314 5 Singular Integrals of Convolution Type

5.1.1 Definition and Basic Properties of the Hilbert Transform

There are several equivalent ways to introduce the Hilbert transform; in this ex-

position we first define it as a convolution operator with a certain principal value

distribution, but we later discuss other equivalent definitions.

We begin by defining a distribution W0 in S ′(R) as follows:

⟨W0,ϕ

⟩=

1

πlimε→0

∫

ε≤|x|≤1

ϕ(x)

xdx+

1

π

∫

|x|≥1

ϕ(x)

xdx , (5.1.1)

for ϕ in S (R). The function 1/x integrated over [−1,−ε]⋃[ε ,1] has mean value

zero, and we may replace ϕ(x) by ϕ(x)−ϕ(0) in the first integral in (5.1.1). Since

(ϕ(x)−ϕ(0))x−1 is controlled by ‖ϕ ′‖L∞ , it follows that the limit in (5.1.1) exists.

To see that W0 is indeed in S ′(R), we note that the estimate

∣∣⟨W0,ϕ⟩∣∣≤ 2

π

∥∥ϕ ′∥∥

L∞+

2

πsupx∈R

|xϕ(x)| (5.1.2)

is valid. This says that W0 ∈S ′(R).

Definition 5.1.1. The truncated Hilbert transform (at height ε) of a function f in

Lp(R), 1≤ p < ∞, is defined by

H(ε)( f )(x) =1

π

∫

|y|≥ε

f (x− y)

ydy =

1

π

∫

|x−y|≥ε

f (y)

x− ydy . (5.1.3)

The Hilbert transform of ϕ ∈S (R) is defined by

H(ϕ)(x) = (W0 ∗ϕ)(x) = limε→0

H(ε)(ϕ)(x) . (5.1.4)

Observe that H(ε)( f ) is well defined for all f ∈ Lp, 1≤ p <∞. This follows from

Holder’s inequality, since 1/x is integrable to the power p′ on the set |x| ≥ ε .

For Schwartz functions ϕ , the integral

∫ +∞

−∞

ϕ(x− y)

ydy

may not converge absolutely for any real number x, but is defined as a limit of the

absolutely convergent integrals

∫

|y|≥ε

ϕ(x− y)

ydy ,

as ε → 0. Such limits are called principal value integrals and are denoted by the

letters p.v. Using this notation, the Hilbert transform of a Schwartz function ϕ is

H(ϕ)(x) =1

πp.v.

∫ +∞

−∞

ϕ(x− y)

ydy =

1

πp.v.

∫ +∞

−∞

ϕ(y)

x− ydy . (5.1.5)

5.1 The Hilbert Transform and the Riesz Transforms 315

Remark 5.1.2. We extend the definition of the Hilbert transform to a bigger class

of functions. Suppose that f is an integrable function on R that satisfies a Holder

condition near every point x; that is, for any x ∈ R there are Cx > 0 and εx > 0 such

that

| f (x)− f (y)| ≤Cx|x− y|εx

whenever |y− x|< δx. Then we write

H(ε)( f )(x) =1

π

∫

ε<|x−y|<δx

f (y)

x− ydy+

1

π

∫

|x−y|≥δx

f (y)

x− ydy

=1

π

∫

ε<|x−y|<δx

f (y)− f (x)

x− ydy+

1

π

∫

|x−y|≥δx

f (y)

x− ydy .

Both integrals converge absolutely; hence the limit of H(ε)( f )(x) exists as ε → 0.

Fig. 5.1 The graph of the

function H(χE) when E is

a union of three disjoint

intervals J1∪ J2∪ J3.

Example 5.1.3. For the characteristic function χ[a,b] of an interval [a,b] we show

that

H(χ[a,b])(x) =1

πlog|x−a||x−b| . (5.1.6)

Let us verify this identity. Pick ε < min(|x− a|, |x− b|). To show (5.1.6) consider

the three cases 0 < x− b, x− a < 0, and x− b < 0 < x− a. In the first two cases,

(5.1.6) follows immediately. In the third case we have

H(χ[a,b])(x) =1

πlimε→0

(log|x−a|ε

+ logε

|x−b|

), (5.1.7)

which yields (5.1.6). Observe that the cancellation of ε in (5.1.7) reflects the fact

that 1/x has integral zero on symmetric intervals ε < |x|< c. Note that H(χ[a,b])(x)

blows up logarithmically in x near the points a and b and decays like |x|−1 as x→∞.

See Figure 5.1.

Example 5.1.4. Let log+ x = logx when x≥ 1 and zero otherwise. Observe that the

calculation in the previous example actually gives

H(ε)(χ[a,b])(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

1

πlog+

|x−a|max(ε , |x−b|) when x > b,

− 1

πlog+

|x−b|max(ε , |x−a|) when x < a,

1

πlog+

|x−a|ε

− 1

πlog+

|x−b|ε

when a < x < b.

We now give an alternative characterization of the Hilbert transform using the

Fourier transform. To achieve this we need to compute the Fourier transform of the

distribution W0 defined in (5.1.1). Fix a Schwartz function ϕ on R. Then

⟨W0,ϕ

⟩=

⟨W0, ϕ

⟩(5.1.8)

=1

πlimε→0

∫

|ξ |≥εϕ(ξ )

dξ

ξ

=1

πlimε→0

∫

1ε≥|ξ |≥ε

∫

Rϕ(x)e−2πixξ dx

dξ

ξ

= limε→0

∫

Rϕ(x)

[1

π

∫

1ε≥|ξ |≥ε

e−2πixξ dξ

ξ

]dx

= limε→0

∫

Rϕ(x)

[−i

π

∫

1ε≥|ξ |≥ε

sin(2πxξ )dξ

ξ

]dx

= limε→0

∫

Rϕ(x)

[(−i

πsgnx

)∫

12πε≥|ξ |≥ ε

2π

sin(|x|ξ ) dξ

ξ

]dx . (5.1.9)

Here we used the signum function

sgnx =

⎧⎪⎨⎪⎩

+1 when x > 0,

0 when x = 0,

−1 when x < 0.

(5.1.10)

Using the results (a) and (b) in Exercise 5.1.1 we obtain that the integrals inside the

square brackets in (5.1.9) are uniformly bounded by 8 and converge to 2π2= π as

ε → 0, whenever x = 0. These observations make possible the use of the Lebesgue

dominated convergence theorem that allows the passage of the limit inside the inte-

gral in (5.1.9). We obtain that

⟨W0,ϕ

⟩=

∫

Rϕ(x)(−isgn(x))dx . (5.1.11)

This implies that

W0(ξ ) =−isgnξ . (5.1.12)

In particular, identity (5.1.12) says that W0 is a (bounded) function.

We now use identity (5.1.12) to write

H( f )(x) =(

f (ξ )(−isgnξ ))∨(x) . (5.1.13)

This formula can be used to give an alternative definition of the Hilbert transform.

An immediate consequence of (5.1.13) is that

∥∥H( f )∥∥

L2 =∥∥ f

∥∥L2 , (5.1.14)

that is, H is an isometry on L2(R). Moreover, H satisfies

H2 = HH =−I , (5.1.15)

where I is the identity operator. Equation (5.1.15) is a simple consequence of the

fact that (−isgnξ )2 =−1. The adjoint operator H∗ of H is uniquely defined via the

identity

⟨f |H(g)

⟩=

∫

Rf H(g)dx =

∫

RH∗( f ) gdx =

⟨H∗( f ) |g

⟩,

and we can easily obtain that H∗ has multiplier −isgnξ = isgnξ . We conclude that

H∗ =−H. Likewise, we obtain Ht =−H.

5.1.2 Connections with Analytic Functions

We now investigate connections of the Hilbert transform with the Poisson kernel.

Recall the definition of the Poisson kernel Py given in Example 1.2.17. Then for a

real-valued function f in Lp(R), 1≤ p < ∞, we have

(Py ∗ f )(x) =y

π

∫ +∞

−∞

f (t)

(x− t)2 + y2dt , (5.1.16)

and the integral in (5.1.16) converges absolutely by Holder’s inequality, since the

function t → ((x− t)2 + y2)−1 is in Lp′(R) whenever y > 0.

Let Re z and Im z denote the real and imaginary parts of a complex number z.

Observe that

(Py ∗ f )(x) = Re

(i

π

∫ +∞

−∞

f (t)

x− t + iydt

)= Re

(i

π

∫ +∞

−∞

f (t)

z− tdt

),

where z = x+ iy. The function

Ff (z) =i

π

∫ +∞

−∞

f (t)

z− tdt

defined on

R2+ = z = x+ iy : y > 0

is analytic, since its ∂/∂ z derivative is zero. The real part of Ff (x+ iy) is (Py ∗ f )(x).The imaginary part of Ff (x+ iy) is

Im

(i

π

∫ +∞

−∞

f (t)

x− t + iydt

)=

1

π

∫ +∞

−∞

f (t)(x− t)

(x− t)2 + y2dt = ( f ∗Qy)(x) ,

where Qy is called the conjugate Poisson kernel and is given by

Qy(x) =1

π

x

x2 + y2. (5.1.17)

The function u f + iv f is analytic and thus u f (x+ iy) = ( f ∗Py)(x) and v f (x+ iy) =( f ∗Qy)(x) are conjugate harmonic functions. Since the family Py, y > 0, is an ap-

proximate identity, it follows from Theorem 1.2.19 that Py∗ f → f in Lp(R) as y→ 0.

The following question therefore arises: What is the limit of f ∗Qy as y→ 0? The

next result addresses this question.

Theorem 5.1.5. Let 1≤ p < ∞. For any f ∈ Lp(R) we have

f ∗Qε −H(ε)( f )→ 0 (5.1.18)

in Lp and almost everywhere as ε → 0. Moreover, for ϕ in S (R) we have

Fϕ(x+ iy) =i

π

∫ +∞

−∞

ϕ(t)

x+ iy− tdt → ϕ(x)+ iH(ϕ)(x) (5.1.19)

as y→ 0+ for all x ∈ R.

Proof. We see that

(Qε ∗ f )(x)− 1

π

∫

|t|≥ε

f (x− t)

tdt =

1

π( f ∗ψε)(x),

where ψε(x) = ε−1ψ(ε−1x) and

ψ(t) =

t

t2+1− 1

twhen |t| ≥ 1,

tt2+1

when |t|< 1.(5.1.20)

Note that ψ is integrable over the line and has integral zero. Furthermore, the inte-

grable function

Ψ(t) =

1

t2+1when |t| ≥ 1,

1 when |t|< 1,(5.1.21)

is a radially decreasing majorant of ψ , i.e., it is even, decreasing on [0,∞), and

satisfies |ψ| ≤Ψ . It follows from Theorem 1.2.21 (with a = 0) that f ∗ψε → 0 in

Lp. Also Corollary 2.1.19 (with a = 0) implies that f ∗ψε → 0 almost everywhere

as ε → 0.

Assertion (5.1.19) is a consequence of (5.1.18), the discussion preceding Theorem

5.1.5, and the observation that H(ε)(ϕ) converges to H(ϕ) pointwise everywhere as

ε → 0.

Remark 5.1.6. We will show later that for f ∈ Lp(R), 1 ≤ p < ∞, the expressions

H(ε)( f ) converge a.e. (and also in Lp when p > 1) to a function H( f ). This will

be a consequence of Theorem 5.1.12 (or Corollary 5.3.6 when p = 1), combined

with Theorem 2.1.14 and the observation that for Schwartz functions ϕ , H(ε)(ϕ)converge to H(ϕ) as ε → 0. The linear operator H defined in this way extends the

Hilbert transform H initially defined on Schwartz functions and will still be denoted

by H. Thus for f ∈ Lp(R), 1≤ p < ∞, one has

limε→0

f ∗Qε = H( f ) a.e.

This convergence is also valid in Lp in view of the preceding observations and The-

orem 5.1.5.

5.1.3 Lp Boundedness of the Hilbert Transform

As a consequence of the result in Exercise 5.1.4 and of the fact that

x≤ 12(ex− e−x) , x≥ 0,

we obtain that

|x : |H(χE)(x)|> α| ≤ 2

π

|E|α

, α > 0, (5.1.22)

for all subsets E of the real line of finite measure. Theorem 1.4.19 with p0 = q0 = 1

and p1 = q1 = 2 now implies that H is bounded on Lp for 1 < p < 2. Duality gives

that H∗ =−H is bounded on Lp for 2 < p < ∞ and hence so is H.

We give another proof of the boundedness of the Hilbert transform H on Lp(R),which has the advantage that it gives the best possible constant in the resulting norm

inequality when p is a power of 2.

Theorem 5.1.7. For all 1 < p < ∞, there exists a positive constant Cp such that

∥∥H( f )∥∥

Lp ≤Cp

∥∥ f∥∥

Lp

for all f in S (R). Moreover, the constant Cp satisfies Cp ≤ 2p for 2 ≤ p < ∞ and

Cp ≤ 2p/(p−1) for 1 < p≤ 2. Therefore, the Hilbert transform H admits an exten-

sion to a bounded operator on Lp(R) when 1 < p < ∞.

Proof. The proof we give is based on the interesting identity

H( f )2 = f 2 +2H( f H( f )), (5.1.23)

which is valid whenever f is a real-valued Schwartz function. We prove (5.1.23) in

two different ways. First we consider the analytic function

Ff (z) =i

π

∫

R

f (t)

z− tdt

defined on the upper half space. We compute its square. Fix z ∈C with Rez > 0 and

f a real-valued Schwartz function. Then for ε > 0 we have

Ff (z)2 =

(i

π

)2 ∫

R

∫

R

f (t) f (s)

(z− t)(z− s)dtds

=

(i

π

)2∫

R

∫

|t−s|>ε

f (t) f (s)

t− s

(1

z− t− 1

z− s

)dtds− 1

π2

∫∫

|t−s|≤ε

f (t) f (s)dtds

(z− t)(z− s)

=

(i

π

)2∫

Rf (t)

∫

|t−s|>ε

f (s)

t− sds

dt

z− t−

(i

π

)2 ∫

Rf (s)

∫

|t−s|>ε

f (t)

t− sdt

ds

z− s

− 1

π2

∫∫

|t−s|≤ε

f (t) f (s)dtds

(z− t)(z− s).

Letting ε → 0 and passing the limit inside the integral by the Lebesgue dominated

convergence theorem, we deduce

Ff (z)2 = i

i

π

∫

R

2 f (t)H( f )(t)

z− tdt . (5.1.24)

We now let Imz→ 0+ in (5.1.24) and use (5.1.19) in Theorem 5.1.5. We obtain

f 2−H( f )2 + i2 f H( f ) =(

f + iH( f ))2

= i(

2 f H( f )+ iH(2 f H( f )

)),

and equating the real parts we deduce (5.1.23).

To give an alternative proof of (5.1.23) we take Fourier transforms. Let

m(ξ ) =−isgnξ

be the symbol of the Hilbert transform. We have

f 2(ξ )+2[H( f H( f ))](ξ )= ( f ∗ f )(ξ )+2m(ξ )( f ∗ H( f ))(ξ )

=∫

Rf (η) f (ξ −η)dη+2m(ξ )

∫

Rf (η) f (ξ −η)m(η)dη (5.1.25)

=∫

Rf (η) f (ξ −η)dη+2m(ξ )

∫

Rf (η) f (ξ −η)m(ξ −η)dη . (5.1.26)

Averaging (5.1.25) and (5.1.26) we obtain

f 2(ξ )+2[H( f H( f ))] (ξ ) =∫

Rf (η) f (ξ −η)

[1+m(ξ )

(m(η)+m(ξ −η)

)]dη .

But the last displayed expression is equal to

∫

Rf (η) f (ξ −η)m(η)m(ξ −η)dη = (H( f )∗ H( f ))(ξ )

in view of the identity

m(η)m(ξ −η) = 1+m(ξ )m(η)+m(ξ )m(ξ −η),

which is valid for all (ξ ,η) ∈ R2 \(0,0) for the function m(ξ ) =−isgnξ .

Having established (5.1.23), we can easily obtain Lp bounds for H when p = 2k

is a power of 2. We already know that H is bounded on Lp with norm one when

p = 2k and k = 1. Suppose that H is bounded on Lp with bound cp for p = 2k for

some k ∈ Z+. Then for a nonzero real-valued function f in C ∞0 we have

∥∥H( f )∥∥

L2p =∥∥H( f )2

∥∥ 12

Lp ≤(∥∥ f 2

∥∥Lp +

∥∥2H( f H( f ))∥∥

Lp

) 12

≤(∥∥ f

∥∥2

L2p +2cp

∥∥ f H( f )∥∥

Lp

) 12

≤(∥∥ f

∥∥2

L2p +2cp

∥∥ f∥∥

L2p

∥∥H( f )∥∥

L2p

) 12 .

Since ‖H( f )‖L2p < ∞, we obtain that

(∥∥H( f )∥∥

L2p∥∥ f∥∥

L2p

)2

−2cp

∥∥H( f )∥∥

L2p∥∥ f∥∥

L2p

−1≤ 0.

If follows that ∥∥H( f )∥∥

L2p∥∥ f∥∥

L2p

≤ cp +√

c2p +1 ,

and from this we conclude that H is bounded on L2p with bound

c2p ≤ cp +√

c2p +1 . (5.1.27)

This completes the induction. We have proved that H maps Lp to Lp when p = 2k,

k = 1,2, . . . . Interpolation now gives that H maps Lp to Lp for all p ≥ 2. Since

H∗ =−H, duality gives that H is also bounded on Lp for 1 < p≤ 2.

The previous proof of the boundedness of the Hilbert transform provides us with

some useful information about the norm of this operator on Lp(R). Let us begin

with the identity

cotx

2= cotx+

√1+ cot2 x,

valid for 0 < x < π2

. If cp ≤ cot π2p

, then (5.1.27) gives that

c2p ≤ cp +√

c2p +1≤ cot

π

2p+

√1+ cot2

π

2p= cot

π

2 ·2p,

and since 1 = cot π4= cot π

2·2 , we obtain by induction that the numbers cot π2p

are

indeed bounds for the norm of H on Lp when p = 2k, k = 1,2, . . . . Duality now

gives that the numbers cot π2p′ = tan π

2pare bounds for the norm of H on Lp when

p = 2k

2k−1, k = 1,2, . . . . These bounds allow us to derive good estimates for the norm

‖H‖Lp→Lp as p→ 1 and p→ ∞. Indeed, since cot π2p≤ p when p ≥ 2, the Riesz–

Thorin interpolation theorem gives that ‖H‖Lp→Lp ≤ 2p for 2 ≤ p < ∞ and by du-

ality ‖H‖Lp→Lp ≤ 2pp−1

for 1 < p ≤ 2. This completes the proof which is worth

comparing with that of Theorem 4.1.7.

Remark 5.1.8. The numbers cot π2p

for 2 ≤ p < ∞ and tan π2p

for 1 < p ≤ 2 are

indeed equal to the norms of the Hilbert transform H on Lp(R). This requires a

more delicate argument; see Exercise 5.1.12.

Remark 5.1.9. We may wonder what happens when p = 1 or p = ∞. The Hilbert

transform of χ[a,b] computed in Example 5.1.3 is easily seen to be unbounded and not

integrable, since it behaves like 1/|x| as x→ ∞. This behavior near infinity suggests

that the Hilbert transform may map L1 to L1,∞. This is indeed the case, but this will

not be shown until Section 5.3.

We now introduce the maximal Hilbert transform.

Definition 5.1.10. The maximal Hilbert transform is the operator

H(∗)( f )(x) = supε>0

∣∣∣H(ε)( f )(x)∣∣∣ (5.1.28)

defined for all f in Lp, 1≤ p<∞. For such f , H(ε)( f ) is well defined as a convergent

integral by Holder’s inequality. Hence H(∗)( f ) makes sense for f ∈ Lp(R), although

for some values of x, H(∗)( f )(x) may be infinite.

Example 5.1.11. Using the result of Example 5.1.4, we obtain that

H(∗)(χ[a,b])(x) =1

π

∣∣∣∣log|x−a||x−b|

∣∣∣∣ . (5.1.29)

We see that in general, H(∗)( f )(x) = |H( f )(x)| by taking f to be the characteristic

function of the union of two disjoint closed intervals.

The definition of H gives that H(ε)( f ) converges pointwise to H( f ) whenever f

lies in C ∞0 (R). If we have the estimate ‖H(∗)( f )‖Lp ≤Cp‖ f‖Lp for f ∈ Lp(R), The-

orem 2.1.14 yields that H(ε)( f ) converges to H( f ) a.e. as ε→ 0 for any f ∈ Lp. This

almost everywhere limit provides a way to describe H( f ) for general f ∈ Lp(R).Note that Theorem 5.1.7 implies only that H has a (unique) bounded extension on

Lp, but it does not provide a way to describe H( f ) when f is a general Lp function.

The next theorem is a simple consequence of these ideas.

Theorem 5.1.12. There exists a constant C such that for all 1 < p < ∞ we have

∥∥H(∗)( f )∥∥

Lp ≤C max(

p,(p−1)−2)∥∥ f

∥∥Lp . (5.1.30)

Moreover, for all f in Lp(R), H(ε)( f ) converges to H( f ) a.e. and in Lp.

Proof. Another proof of this theorem is given in Theorem 4.2.4 in [131] in which

the asserted bound is improved.

Recall the kernels Pε and Qε defined in (5.1.16) and (5.1.17). Fix 1 0 , (5.1.31)

holds whenever f is an Lp function. Then we have

H(ε)( f ) = H(ε)( f )− f ∗Qε +H( f )∗Pε . (5.1.32)

Using the identity

H(ε)( f )(x)− ( f ∗Qε)(x) =−1

π

∫

Rf (x− t)ψε(t)dt , (5.1.33)

where ψ is as in (5.1.20), and applying Corollary 2.1.12, we obtain the estimate

supε>0

|H(ε)( f )(x)− ( f ∗Qε)(x)| ≤1

π

∥∥Ψ∥∥

L1 M( f )(x) , (5.1.34)

whereΨ is as in (5.1.21) and M is the Hardy–Littlewood maximal function. In view

of (5.1.32) and (5.1.34), we obtain for f ∈ Lp(Rn) that

|H(∗)( f )(x)| ≤∥∥Ψ

∥∥L1 M( f )(x)+M(H( f ))(x) . (5.1.35)

It follows immediately from (5.1.35) that H(∗) is Lp bounded with norm at most

C max(

p,(p−1)−2).


We now turn to the proof of (5.1.31). It suffices to prove (5.1.31) for Schwartz

functions since, given f ∈ Lp there is a sequence φ j ∈S such that ‖ f −φ j‖Lp → 0

as j → ∞ and Pε , Qε lie in Lp′ . Taking Fourier transforms, we see that (5.1.31) is a

consequence of the identity

((−isgnξ )e−2π|ξ |)∨(x) = 1

π

x

x2 +1. (5.1.36)

To prove (5.1.36) we write

((−isgnξ )e−2π|ξ |)∨(x) =

∫ +∞

−∞e−2π|ξ |(−isgnξ )e2πixξ dξ

= 2

∫ ∞

0e−2πξ sin(2πxξ )dξ

=1

π

∫ ∞

0e−ξ sin(xξ )dξ (5.1.37)

=1

π

∫ ∞

0(e−ξ )′′ sin(xξ )dξ

= − x

π

∫ ∞

0(e−ξ )′ cos(xξ )dξ

= − x

π

[−1+ x

∫ ∞

0e−ξ sin(xξ )dξ

](5.1.38)

and we equate (5.1.38) and (5.1.37).

The statement in the theorem about the almost everywhere convergence of

H(ε)( f ) to H( f ) is a consequence of (5.1.30), of the fact that the alleged conver-

gence holds for Schwartz functions, and of Theorem 2.1.14. Finally, the Lp conver-

gence follows from the almost everywhere convergence and the Lebesgue dominated

convergence theorem in view of the validity of (5.1.35).

5.1.4 The Riesz Transforms

We now study an n-dimensional analogue of the Hilbert transform. It turns out that

there exist n operators in Rn, called the Riesz transforms, with properties analogous

to those of the Hilbert transform on R.

To define the Riesz transforms, we first introduce tempered distributions Wj on

Rn, for 1≤ j ≤ n, as follows. For ϕ ∈S (Rn), let

⟨Wj,ϕ

⟩=Γ ( n+1

2)

πn+1

2

limε→0

∫

|y|≥ε

y j

|y|n+1ϕ(y)dy.

One should check that indeed Wj ∈S ′(Rn). Observe that the normalization of Wj

is similar to that of the Poisson kernel.

Definition 5.1.13. For 1≤ j≤ n, the jth Riesz transform of f is given by convolution

with the distribution Wj, that is,

R j( f )(x) = ( f ∗Wj)(x) =Γ ( n+1

2)

πn+1

2

p.v.

∫

Rn

x j− y j

|x− y|n+1f (y)dy , (5.1.39)

for all f ∈S (Rn). Definition 5.1.13 makes sense for any integrable function f that

has the property that for all x there exist Cx > 0, εx > 0, and δx > 0 such that for

y satisfying |y− x| < δx we have | f (x)− f (y)| ≤ Cx|x− y|εx . The principal value

integral in (5.1.39) is as in Definition 5.1.1.

We now give a characterization of R j using the Fourier transform. For this we

need to compute the Fourier transform of Wj.

Proposition 5.1.14. The jth Riesz transform R j is given on the Fourier transform

side by multiplication by the function −iξ j/|ξ |. That is, for any f in S (Rn) we

have

R j( f )(x) =(− iξ j

|ξ | f (ξ ))∨

(x) . (5.1.40)

Proof. The proof is essentially a reprise of the corresponding proof for the Hilbert

transform, but it involves a few technical difficulties. Fix a Schwartz function ϕ on

Rn. Then for 1≤ j ≤ n we have

⟨Wj,ϕ

⟩=

⟨Wj, ϕ

⟩(5.1.41)

=Γ ( n+1

2)

πn+1

2

limε→0

∫

|ξ |≥εϕ(ξ )

ξ j

|ξ |n+1dξ

=Γ ( n+1

2)

πn+1

2

limε→0

∫

1ε≥|ξ |≥ε

∫

Rnϕ(x)e−2πix·ξ dx

ξ j

|ξ |n+1dξ

= limε→0

∫

Rnϕ(x)

[Γ ( n+1

2)

πn+1

2

∫

1ε≥|ξ |≥ε

e−2πix·ξ ξ j

|ξ |n+1dξ

]dx

= limε→0

∫

Rnϕ(x)

[Γ ( n+1

2)

πn+1

2

∫

Sn−1

∫

ε≤r≤ 1ε

e−2πirx·θ r

rn+1rn−1drθ jdθ

]dx

=∫

Rnϕ(x)

[−iΓ ( n+1

2)

πn+1

2

∫

Sn−1

∫ ∞

0sin(2πrx ·θ) dr

rθ j dθ

]dx

=∫

Rnϕ(x)

[−iπ

2

Γ ( n+12)

πn+1

2

∫

Sn−1sgn(x ·θ)θ j dθ

]dx

=∫

Rn−iϕ(x)

x j

|x|dx ,

where in the penultimate equality we used the identity∫ ∞

0sin t

tdt = π

2, for which we

refer to Exercise 5.1.1, while in the last equality we used the identity

−iπ

2

Γ ( n+12)

πn+1

2

∫

Sn−1sgn(x ·θ)θ j dθ =−i

x j

|x| , (5.1.42)

which needs to be established. The passage of the limit inside the integral in the

previous calculation is a consequence of the Lebesgue dominated convergence the-

orem, which is justified from the fact that

∣∣∣∣∫ 1/ε

ε

sin(2πrθ)

rdr

∣∣∣∣≤ 4 (5.1.43)

for all ε > 0. For a proof of (5.1.43) we again refer to Exercise 5.1.1.

It remains to establish (5.1.42). Let us recall that O(n) is the set of all orthogonal

n× n matrices with real entries. An invertible matrix A is called orthogonal if its

transpose At is equal to its inverse A−1, that is, AAt = AtA = I.

Lemma 5.1.15. The following identity is valid for all ξ ∈ Rn \0:∫

Sn−1sgn(ξ ·θ)θ j dθ =

2πn−1

2

Γ ( n+12)

ξ j

|ξ | . (5.1.44)

Therefore (5.1.42) holds.

Proof. We begin with the identity

∫

Sn−1sgn(θk)θ j dθ =

⎧⎪⎪⎨⎪⎪⎩

0 if k = j,

∫

Sn−1|θ j|dθ if k = j,

(5.1.45)

which can be proved by noting that for k = j, sgn(θk) has a constant sign on the

hemispheres θk > 0 and θk < 0, on either of which the function θ → θ j has integral

zero.

It suffices to prove (5.1.44) for a unit vector ξ . Given ξ ∈ Sn−1, pick an orthogonal

n×n matrix A = (akl)k,l such that Ae j = ξ . Then the jth column of the matrix A is

the vector (ξ1,ξ2, . . . ,ξn)t . We have

∫

Sn−1sgn(ξ ·θ)θ j dθ =

∫

Sn−1sgn(Ae j ·θ)θ j dθ

=∫

Sn−1sgn(e j ·Atθ)(AAtθ) j dθ

=∫

Sn−1sgn(e j ·θ)(Aθ) j dθ

=∫

Sn−1sgn(θ j)(a j1θ1 + · · ·+ξ jθ j + · · ·+a jnθn)dθ


= ξ j

∫

Sn−1sgn(θ j)θ j dθ + ∑

1≤m= j≤n

0

=ξ j

|ξ |

∫

Sn−1|θ j|dθ .

Next, for all j ∈ 1,2, . . . ,n, we compute the value of the integral

∫

Sn−1|θ j|dθ =

∫

Sn−1|θ1|dθ ,

which is obviously independent of j by symmetry. In view of the result of Appendix

D.2, we write

∫

Sn−1|θ1|dθ =

∫ 1

−1|s|

∫√

1−s2 Sn−2dϕ

ds

(1− s2)12

= ωn−2

∫ 1

−1|s|(1− s2)

n−32 ds

= ωn−2

∫ 1

0u

n−32 du

=2ωn−2

n−1

=2π

n−12

Γ ( n−12) n−1

2

=2π

n−12

Γ ( n+12),

having used the expression for ωn−2 in Appendix A.3. This proves (5.1.44). The

proof of the lemma and hence that of Proposition 5.1.14 is complete.

Proposition 5.1.16. The Riesz transforms satisfy

−I =n

∑j=1

R2j , on L2(Rn), (5.1.46)

where I is the identity operator.

Proof. Use the Fourier transform and the identity ∑nj=1(−iξ j/|ξ |)2 =−1 to obtain

that ∑nj=1 R2

j( f ) =− f for any f in L2(Rn).

We can express the mixed derivatives of Schwartz function in terms of its Lapla-

cian using the Riesz transforms.


Proposition 5.1.17. For ϕ in S (Rn) and 1≤ j,k ≤ n we have

∂ j∂kϕ(x) =−R jRk∆ϕ(x) (5.1.47)

for all x ∈ Rn.

Proof. We verify the claimed identity by taking Fourier transforms. We have

(∂ j∂kϕ

)(ξ ) = (2πiξ j)(2πiξk)ϕ(ξ )

= −(− iξ j

|ξ |)(− iξk

|ξ |)(−4π2|ξ |2)ϕ(ξ )

= −(R jRk∆ϕ

)(ξ )

and taking the inverse Fourier transform, identity (5.1.47) follows.

Next we discuss a use of the Riesz transforms to partial differential equations.

Example 5.1.18. Suppose that f is a given function in L2(Rn) and that u is a tem-

pered distribution on Rn that solves Laplace’s equation

∆u = f . (5.1.48)

We express all second-order derivatives of u in terms of the Riesz transforms of f .

To solve equation (5.1.48) we first show that the tempered distribution

(∂ j∂ku+R jRk( f )

)

is supported at 0. In view of Proposition 2.4.1, this implies that

∂ j∂ku =−R jRk( f )+P

where P is a polynomial of n variables (that depends on j and k) and provides a way

to express the mixed partials of u in terms of the Riesz transforms of f .

To verify that(∂ j∂ku+R jRk( f )

) is supported at 0, we fix a Schwartz function

ψ whose support does not contain the origin. Then ψ vanishes in a neighborhood

of zero and we can pick C ∞ function η which vanishes in a smaller neighborhood

of zero and is equal to 1 on the support of ψ . We define

ζ (ξ ) =−η(ξ )(− iξ j

|ξ |)(− iξk

|ξ |)

and we notice that ζ is a bounded C ∞ function and so are all of its derivatives; also

η(ξ )(2πiξ j)(2πiξk) = ζ (ξ )(−4π2|ξ |2) .

Taking the Fourier transform of both sides of (5.1.48) we obtain

(−4π2|ξ |2) u(ξ ) = f (ξ )

and multiplying by ζ (which is allowed since all derivatives of ζ lie in L∞∩C ∞)

ζ (ξ )(−4π2|ξ |2) u = ζ (ξ ) ∆u = ζ (ξ ) f (ξ ) .

It follows that for all 1≤ j,k ≤ n we have

⟨(∂ j∂ku),ψ

⟩=

⟨(2πiξ j)(2πiξk)u,ψ

⟩

=⟨(2πiξ j)(2πiξk)u,ηψ

⟩

=⟨η(ξ )(2πiξ j)(2πiξk)u,ψ

⟩

=⟨ζ (ξ )(−4π2|ξ |2)u,ψ

⟩

=⟨ζ (ξ ) f (ξ ),ψ

⟩

=⟨−η(ξ )

(− iξ j

|ξ |)(− iξk

|ξ |)

f (ξ ),ψ⟩

=⟨−η(ξ )(R jRk( f ))(ξ ),ψ

⟩

= −⟨(R jRk( f )),ηψ

⟩

= −⟨(R jRk( f )),ψ

⟩

and since this holds for all Schwartz functions ψ whose support does not contain

the origin, it follows that(∂ j∂ku+R jRk( f )

) is supported at 0.

Exercises

5.1.1. (a) Show that for all 0 < a 0 define

I(a) =∫ ∞

0

sinx

xe−ax dx

and show that I(a) is continuous at zero. Differentiate in a and look at the behavior

of I(a) as a→ ∞ to obtain the identity

I(a) =π

2− arctan(a) .

Deduce that I(0) = π2

and also derive the following identity used in (5.1.10):

∫ +∞

−∞

sin(bx)

xdx = π sgn(b) .

(c) Argue as in part (b) to prove for a≥ 0 the identity

∫ ∞

0

1− cosx

x2e−ax dx =

π

2− arctan(a)+a log

a√1+a2

.

[Hint: Part (a): Consider the cases b ≤ 1, a ≤ 1 ≤ b, 1 ≤ a. When a ≥ 1, integrate

by parts.]

5.1.2. (a) Let ϕ be a compactly supported C m+1 function on R for some m in

Z+⋃0. Prove that if ϕ(m) is the mth derivative of ϕ , then

|H(ϕ(m))(x)| ≤Cm,ϕ (1+ |x|)−m−1

for some Cm,ϕ > 0.

(b) Let ϕ be a compactly supported C m+1 function on Rn for some m ∈ Z+. Show

that

|R j(∂αϕ)(x)| ≤Cn,m,ϕ (1+ |x|)−n−m

for some Cn,m,ϕ > 0 and all multi-indices α with |α|= m.

(c) Let I be an interval on the line and assume that a function h is equal to 1 on the

left half of I, is equal to −1 on the right half of I, and vanishes outside I. Prove that

for x /∈ 2I we have

|H(h)(x)| ≤ 4|I|2|x− center(I)|−2 .[Hint: Use that when |t| ≤ 1

2we have log(1+ t) = t +R1(t), where |R1(t)| ≤ 2|t|2.

]

5.1.3. (a) Using identity (5.1.13) one may define H( f ) as an element of S ′(R) for

bounded functions f on the line whose Fourier transform vanishes in a neighbor-

hood of the origin. Using this interpretation, prove that

H(eix) = − ieix ,

H(cosx) = sinx ,

H(sinx) = − cosx ,

H(sin(πx)/πx) = (1− cos(πx))/πx .

(b) Show that the operators given by convolution with the smooth function sin(t)/t

and the distribution p.v. cos(t)/t are bounded on Lp(R) whenever 1 < p < ∞.[Hint: Use that the Fourier transform of the distribution eix is δ1/2π .

]

5.1.4. ([347]) Show that the distribution function of the Hilbert transform of the

characteristic function of a measurable subset E of the real line of finite measure is

dH(χE )(α) =4|E|

eπα − e−πα, α > 0 .

[Hint: First take E =

⋃Nj=1(a j,b j), where b j < a j+1. Show that the equation

H(χE)(x) = α has exactly one root ρ j in each open interval (a j,b j) for 1 ≤ j ≤ N

and exactly one root r j in each interval (b j,a j+1) for 1≤ j ≤ N, (aN+1 = ∞). Then

|x ∈ R : H(χE)(x)> α|= ∑Nj=1 r j−∑N

j=1ρ j, and this can be expressed in terms

of ∑Nj=1 a j and ∑N

j=1 b j. Argue similarly for the set x ∈R : H(χE)(x)<−α. For a

general measurable set E, find sets En such that each En is a finite union of intervals

and that χEn → χE in L2. Then H(χEn)→H(χE) in measure; thus H(χEnk)→H(χE)

a.e. for some subsequence nk. The Lebesgue dominated convergence theorem gives

dH(χEnk)→ dH(χE ). See Figure 5.1.

]

5.1.5. Let 1≤ p <∞ and let T be a linear operator defined on the space of Schwartz

functions that commutes with dilations, i.e., T (δ λ f ) = δ λT ( f ) for all f ∈S (Rn)and all λ > 0. (Here δ λ ( f )(x) = f (λx).) Suppose that there exists a constant C > 0

such that for all f ∈S (Rn) with Lp norm one we have

|x : |T ( f )(x)|> 1| ≤C.

Prove that T admits a bounded extension from Lp(Rn) to Lp,∞(Rn) with norm at

most C1/p.[Hint: Try functions of the form λ−n/p f (λ−1x)/‖ f‖Lp with λ > 0.

]

5.1.6. Let ϕ be in S (R). Prove that

limN→∞

p.v.

∫

R

e2πiNx

xϕ(x)dx = ϕ(0)πi,

limN→−∞

p.v.

∫

R

e2πiNx

xϕ(x)dx = −ϕ(0)πi.

5.1.7. Let Tα , α ∈ R, be the operator given by convolution with the distribution

whose Fourier transform is the function

uα(ξ ) = e−πiα sgnξ .

(a) Show that the Tα ’s are isometries on L2(R) that satisfy

(Tα)−1 = T2−α .

(b) Express Tα in terms of the identity operator and the Hilbert transform.

5.1.8. Let Q( j)y be the jth conjugate Poisson kernel of Py defined by

Q( j)y (x) =

Γ ( n+12)

πn+1

2

x j

(|x|2 + y2)n+1

2

.

Prove that

(Q( j)y )∧ (ξ ) =−i

ξ j

|ξ |e−2πy|ξ | .

Conclude that R j(Py) = Q( j)y and that for f in L2(Rn) we have R j( f )∗Py = f ∗Q

( j)y .

These results are analogous to the statements Qy(ξ ) =−isgn(ξ )Py(ξ ), H(Py) =Qy,

and H( f )∗Py = f ∗Qy.

5.1.9. Fix n ≥ 2. Let f0, f1, . . . , fn be in L2(Rn) and, for 0 ≤ j ≤ n, let u j(x,x0) =(Px0

∗ f j)(x) be the Poisson integrals of f j where x = (x1, . . . ,xn) ∈ Rn and x0 > 0.

Show that a necessary and sufficient condition for

f j = R j( f0), j = 1, . . . ,n,

is that the following system of generalized Cauchy-Riemann equations holds:

n

∑j=0

∂u j

∂x j

(x,x0) = 0 ,

∂u j

∂xk

(x,x0) =∂uk

∂x j

(x,x0) , 0≤ j = k ≤ n .

5.1.10. Prove the distributional identity

∂ j|x|−n+1 = (1−n)p.v.x j

|x|n+1.

Then take Fourier transforms of both sides and use Theorem 2.4.6 to obtain another

proof of Proposition 5.1.14.

5.1.11. (a) Prove that if T is a bounded linear operator on L2(R) that commutes

with translations and dilations and anticommutes with the reflection f (x) → f (x) =f (−x), then T is a constant multiple of the Hilbert transform.

(b) Prove that if T is a bounded linear operator on L2(R) that commutes with transla-

tions and dilations and vanishes when applied to functions whose Fourier transform

is supported in [0,∞), then T is a constant multiple of the operator f →(

f χ(−∞,0])∨

.

5.1.12. ([282]) Fix 1 < p≤ 2.

(a) Show that the function G(x,y) = Re (|x|+ iy)p is subharmonic on R2.

(b) Let u(x,y),v(x,y) be real-valued functions on R2 such that u+ iv is a holomor-

phic function of x+ iy. Prove that G(u,v) is a subharmonic function on R2.

(c) Prove that there is a constant Bp such that for all a and b reals we have

|b|p ≤(

tanπ

2p

)p

|a|p−Bp Re (|a|+ ib)p .

(d) Prove that for f in C ∞0 (R) we have

∫

RRe (| f (x)|+ iH( f )(x))p dx≥ 0.

(e) Combine the results in parts (d) and (c) with a = f (x), b = H( f )(x) to obtain

that∥∥H

∥∥Lp→Lp ≤ tan

π

2p.

5.2 Singular Integrals and the Method of Rotations 333

(f) To deduce that this constant is sharp, take π/2p′ < γ < π/2p and let fγ(x) =

(x+1)−1|x+1|2γ/π |x−1|−2γ/π cosγ . Then

H( fγ)(x) =

⎧⎨⎩

1x+1

|x+1||x−1|

2γ/πsinγ when |x|> 1,

− 1x+1

|x+1||x−1|

2γ/πsinγ when |x|< 1.

[Hint: Part (d): Let CR be the circle of radius R centered at (0,R) in R2. Use that

the integral of the subharmonic function G((Py ∗ f )(x),Qy ∗ f )(x)) over CR is at

least 2πRRe (|(PR ∗ f )(0)|+ i(QR ∗ f )(0))p and let R→∞. Part (f): The formula for

H( fγ) is best derived by considering the restriction of the analytic function

F(z) = (z+1)−1

(iz+ i

z−1

)2γ/π

on the real line.]

5.2 Homogeneous Singular Integrals and the Method

of Rotations

So far we have introduced the Hilbert and the Riesz transforms and we have de-

rived the Lp boundedness of the former. The boundedness properties of the Riesz

transforms on Lp spaces are consequences of the results discussed in this section.

5.2.1 Homogeneous Singular and Maximal Singular Integrals

We introduce singular integral operators on Rn that appropriately generalize the

Riesz transforms on Rn. Here is the setup. We fix Ω to be an integrable function of

the unit sphere Sn−1 with mean value zero. Observe that the kernel

KΩ (x) =Ω(x/|x|)|x|n , x = 0, (5.2.1)

is homogeneous of degree −n just like the functions x j/|x|n+1. Since KΩ is not

in L1(Rn), convolution with KΩ cannot be defined as an operation on Schwartz

functions on Rn. For this reason we introduce a distribution WΩ in S ′(Rn) by setting

⟨WΩ ,ϕ

⟩= lim

ε→0

∫

|x|≥εKΩ (x)ϕ(x)dx = lim

ε→0

∫

ε≤|x|≤ε−1KΩ (x)ϕ(x)dx (5.2.2)

for ϕ ∈S (Rn). Using the fact that Ω has mean value zero, we can easily see that

WΩ is a well defined tempered distribution on Rn. Indeed, since KΩ has integral zero

over all annuli centered at the origin, we have

∣∣⟨WΩ ,ϕ⟩∣∣ =

∣∣∣∣limε→0

∫

ε≤|x|≤1

Ω(x/|x|)|x|n (ϕ(x)−ϕ(0))dx+

∫

|x|≥1


∣∣∣∣

≤∥∥∇ϕ

∥∥L∞

∫

|x|≤1

|Ω(x/|x|)||x|n−1

dx+ supy∈Rn

|y| |ϕ(y)|∫

|x|≥1

|Ω(x/|x|)||x|n+1

dx

≤C1

∥∥∇ϕ∥∥

L∞

∥∥Ω∥∥

L1 +C2 ∑|α |≤1

∥∥ϕ(x)xα∥∥

L∞

∥∥Ω∥∥

L1 ,

for suitable C1 and C2, where we used (2.2.2) in the last estimate. Note that the

distribution WΩ coincides with the function KΩ on Rn \0.The Hilbert transform and the Riesz transforms are examples of these general

operators TΩ . For instance, the function Ω(θ) = θπ|θ | =

1π sgnθ defined on the unit

sphere S0 = −1,1 R gives rise to the Hilbert transform, while the function

Ω(θ) =Γ ( n+1

2)

πn+1

2

θ j

|θ |

defined on Sn−1 Rn gives rise to the jth Riesz transform.

Definition 5.2.1. Let Ω be integrable on the sphere Sn−1 with mean value zero. For

0 < ε < N and f ∈⋃1≤p<∞Lp(Rn) we define the truncated singular integral

T(ε ,N)Ω ( f )(x) =

∫

ε≤|y|≤Nf (x− y)

Ω(y/|y|)|y|n dy . (5.2.3)

Note that for f ∈ Lp(Rn) we have

∥∥T(ε ,N)Ω ( f )

∥∥Lp ≤

∥∥Ω∥∥

L1 log(N/ε)∥∥ f

∥∥Lp(Rn)

,

which implies that (5.2.3) is finite a.e. and therefore well defined. We denote by TΩthe singular integral operator whose kernel is the distribution WΩ , that is,

TΩ ( f )(x) = ( f ∗WΩ )(x) = limε→0N→∞

T(ε ,N)Ω ( f )(x) ,

defined for f ∈S (Rn). The associated maximal singular integral is defined by

T(∗∗)Ω ( f ) = sup

0<N<∞sup

0<ε<N

∣∣T (ε ,N)Ω ( f )

∣∣. (5.2.4)

We note that if Ω is bounded, there is no need to use the upper truncations in

the definition of T(ε ,N)Ω given in (5.2.3). In this case the maximal singular integrals

could be defined as

T(∗)Ω ( f ) = sup

ε>0

∣∣T (ε)Ω ( f )

∣∣ , (5.2.5)

where for f ∈ ⋃1≤p<∞Lp(R), ε > 0, and x ∈ Rn, T

(ε)Ω ( f )(x) is defined in terms of

the absolutely convergent integral

T(ε)Ω ( f )(x) =

∫

|y|≥εf (x− y)

Ω(y/|y|)|y|n dy .

To examine the relationship between T(∗)Ω and T

(∗∗)Ω for Ω ∈ L∞(Sn−1), notice that

∣∣∣∣∫


Ω(y/|y|)|y|n dy

∣∣∣∣≤ sup0<N<∞

∣∣T (ε ,N)Ω ( f )(x)

∣∣ . (5.2.6)

Then for f ∈ Lp(Rn), 1 ≤ p < ∞, we let N → ∞ on the left in (5.2.6) and we note

that the limit exists in view of the absolute convergence of the integral. Then we

take the supremum over ε > 0 to deduce that T(∗)Ω is pointwise bounded by T

(∗∗)Ω .

Since T(ε ,N)Ω = T

(ε)Ω −T

(N)Ω , it also follows that T

(∗∗)Ω ≤ 2T

(∗)Ω ; thus T

(∗)Ω and T

(∗∗)Ω

are pointwise comparable whenΩ lies in L∞(Sn−1). This is the case with the Hilbert

transform, that is, H(∗∗) is comparable to H(∗); likewise with the Riesz transforms.

A certain class of multipliers can be realized as singular integral operators of the

kind discussed. Recall from Proposition 2.4.7 that if m is homogeneous of degree 0

and infinitely differentiable on the sphere, then m∨ is given by

m∨ = cδ0 +WΩ ,

for some complex constant c and some smooth Ω on Sn−1 with mean value zero.

Therefore, all convolution operators whose multipliers are homogeneous of degree

zero smooth functions on Sn−1 can be realized as a constant multiple of the identity

plus an operator of the form TΩ .

Example 5.2.2. Let P(ξ ) = ∑|α |=k bαξα be a homogeneous polynomial of degree

k in Rn that vanishes only at the origin. Let α be a multi-index of order k. Then the

function

m(ξ ) =ξα

P(ξ )(5.2.7)

is infinitely differentiable on the sphere and homogeneous of degree zero. The oper-

ator given by multiplication on the Fourier transform by m(ξ ) is a constant multiple

of the identity plus an operator given by convolution with a distribution of the form

WΩ for some Ω in C ∞(Sn−1) with mean value zero. In this section we establish

the Lp boundedness of such operators when Ω has appropriate smoothness on the

sphere. This, in particular, implies that m(ξ ) defined by (5.2.7) lies in the space

Mp(Rn), defined in Section 2.5, for 1 < p < ∞.

5.2.2 L2 Boundedness of Homogeneous Singular Integrals

Next we would like to compute the Fourier transform of WΩ . This provides infor-

mation as to whether the operator given by convolution with KΩ is L2 bounded. We

have the following result.

Proposition 5.2.3. Let n ≥ 2 and Ω ∈ L1(Sn−1) have mean value zero. Then the

Fourier transform of WΩ is a (finite a.e.) function given by the formula

WΩ (ξ ) =∫

Sn−1Ω(θ)

(log

1

|ξ ·θ | −iπ

2sgn (ξ ·θ)

)dθ . (5.2.8)

Remark 5.2.4. We need to show that the function of ξ on the right in (5.2.8) is well

defined and finite for almost all ξ in Rn. Write ξ = |ξ |ξ ′ where ξ ′ ∈ Sn−1 and notice

that

log1

|ξ ·θ | = log1

|ξ | + log1

|ξ ′ ·θ | .

Since Ω has mean value zero, the term log 1|ξ | multiplied by Ω(θ) vanishes when

integrated over the sphere.

We need to show that

∫

Sn−1|Ω(θ)| log

1

|ξ ′ ·θ | dθ < ∞ (5.2.9)

for almost all ξ ′ ∈ Sn−1. Integrate (5.2.9) over ξ ′ ∈ Sn−1 and apply Fubini’s theorem

to obtain

∫

Sn−1|Ω(θ)|

∫

Sn−1log

1

|ξ ′ ·θ | dξ ′ dθ

=∫

Sn−1|Ω(θ)|

∫

Sn−1log

1

|ξ1|dξ dθ

= ωn−2

∫

Sn−1|Ω(θ)|

∫ +1

−1

(log

1

|s|

)(1− s2)

n−32 dsdθ

=Cn

∥∥Ω∥∥

L1(Sn−1)< ∞ ,

since we are assuming that n ≥ 2. (The second-to-last identity follows from the

identity in Appendix D.2.) We conclude that (5.2.9) holds for almost all ξ ′ ∈ Sn−1.

Since the function of ξ on the right in (5.2.8) is homogeneous of degree zero, it

follows that it is a locally integrable function on Rn.

Before we return to the proof of Proposition 5.2.3, we discuss the following

lemma:

Lemma 5.2.5. Let a be a nonzero real number. Then for 0 < ε < N < ∞ we have

limε→0N→∞

∫ N

ε

cos(ra)− cos(r)

rdr = log

1

|a| , (5.2.10)


∣∣∣∣∫ N

ε

cos(ra)− cos(r)

rdr

∣∣∣∣ ≤ 2

∣∣∣ log1

|a|∣∣∣ for all N > ε > 0 , (5.2.11)

limε→0N→∞

∫ N

ε

e−ira− cos(r)

rdr = log

1

|a| − iπ

2sgn a , (5.2.12)

∣∣∣∣∫ N

ε

e−ira− cos(r)

rdr

∣∣∣∣ ≤ 2

∣∣∣ log1

|a|∣∣∣+4 for all N > ε > 0 . (5.2.13)

Proof. We first prove (5.2.10) and (5.2.11). By the fundamental theorem of calculus

we write

∫ N

ε

cos(ra)− cos(r)

rdr =

∫ N

ε

cos(r|a|)− cos(r)

rdr

= −∫ N

ε

∫ |a|

1sin(tr)dt dr

= −∫ |a|

1

∫ N

εsin(tr)dr dt

= −∫ |a|

1

cos(εt)

tdt +

∫ N|a|

N

cos(t)

tdt ,

and from this expression, we clearly obtain (5.2.11). But the first integral of the

same expression converges to− log |a| as ε→ 0 while the second integral converges

to zero as N → ∞ by an integration by parts. This proves (5.2.10).

To prove (5.2.12) and (5.2.13) we need to know that the expressions

∣∣∣∣∫ N

ε

sin(ra)

rdr

∣∣∣∣=∣∣∣∣∫ N|a|

ε |a|

sin(r)

rdr

∣∣∣∣ (5.2.14)

tend to π2

as ε → 0 and N → ∞ and are bounded by 4. Both statements follow from

Exercise 5.1.1.

Let us now prove Proposition 5.2.3.

Proof. Let us set ξ ′ = ξ/|ξ |. We have the following:

⟨WΩ ,ϕ

⟩=

⟨WΩ , ϕ

⟩

= limε→0

∫

|x|≥ε


= limε→0N→∞

∫

ε≤|x|≤N


= limε→0N→∞

∫

Rnϕ(ξ )

∫

ε≤|x|≤N

Ω(x/|x|)|x|n e−2πix·ξ dx dξ

= limε→0N→∞

∫

Rnϕ(ξ )

∫

Sn−1Ω(θ)

∫

ε≤r≤N

e−2πirθ ·ξ dr

rdθ dξ

= limε→0N→∞

∫

Rnϕ(ξ )

∫

Sn−1Ω(θ)

∫

ε≤r≤N

(e−2πr|ξ |iθ ·ξ ′ − cos(2πr|ξ |)

)dr

rdθ dξ

= limε→0N→∞

∫

Rnϕ(ξ )

∫

Sn−1Ω(θ)

∫

ε2π|ξ |≤r≤ N

2π|ξ |

e−irθ ·ξ ′ − cos(r)

rdr dθ dξ

=

∫

Rnϕ(ξ )

∫

Sn−1Ω(θ)

(log

1

|ξ ′ ·θ | −iπ

2sgn(ξ ·θ)

)dθ dξ ,

where we used the Lebesgue dominated convergence theorem to pass the limit in-

side, Lemma 5.2.5, and Remark 5.2.4. We were able to subtract cos(2πr|ξ |) from

the r integral in the previous calculation, since Ω has mean value zero over the

sphere. Also, the use of the dominated convergence theorem is justified from the

fact that the function

(θ ,ξ ) → |Ω(θ)| |ϕ(ξ )|(

log1

|ξ ′ ·θ | +4)

lies in L1(Sn−1×Rn). Moreover, all the interchanges of integrals are well justified

by Fubini’s theorem.

Corollary 5.2.6. Let Ω ∈ L1(Sn−1) have mean value zero. Then for almost all ξ ′ in

Sn−1 the integral ∫

Sn−1Ω(θ) log

1

|ξ ′ ·θ |dθ (5.2.15)

converges absolutely. Moreover, the associated operator TΩ maps L2(Rn) to itself if

and only if

ess.supξ ′∈Sn−1

∣∣∣∣∫

Sn−1Ω(θ) log

1

|ξ ′ ·θ |dθ∣∣∣∣< ∞ . (5.2.16)

Proof. To obtain the absolute convergence of the integral in (5.2.15) we integrate

over ξ ′ ∈ Sn−1 and we apply Fubini’s theorem. The assertion concerning the bound-

edness of TΩ on L2 is an immediate consequence of Proposition 5.2.3 and Theorem

2.5.10.

There exist functions Ω in L1(Sn−1) with mean value zero such that the ex-

pressions in (5.2.16) are equal to infinity; consequently, not all such Ω give rise

to bounded operators on L2(Rn). Observe, however, that for Ω odd i.e., Ω(−θ) =−Ω(θ) for all θ ∈ Sn−1, (5.2.16) trivially holds, since log 1

|ξ ·θ | is even and its prod-

uct against an odd function must have integral zero over Sn−1. We conclude that

singular integrals TΩ with odd Ω are always L2 bounded.

5.2.3 The Method of Rotations

Having settled the issue of L2 boundedness for singular integrals of the form TΩ with

Ω odd, we turn our attention to their Lp boundedness. A simple procedure called the

method of rotations plays a crucial role in the study of operators TΩ when Ω is an

odd function. This method is based on the use of the directional Hilbert transforms.

Fix a unit vector θ in Rn. For a Schwartz function f on Rn let

Hθ ( f )(x) =1

πp.v.

∫ +∞

−∞f (x− tθ)

dt

t. (5.2.17)

We call Hθ ( f ) the directional Hilbert transform of f in the direction θ . For func-

tions f ∈S (Rn) the integral in (5.2.17) is well defined, since it converges rapidly at

infinity and by subtracting the constant f (x), it also converges near zero.

Likewise, we define the directional maximal Hilbert transforms. For a function

f in⋃

1≤p<∞Lp(Rn) and 0 < ε < N < ∞ we let

H(ε ,N)θ ( f )(x) =

1

π

∫

ε≤|t|≤N

f (x− tθ)dt

t,

H(∗∗)θ ( f )(x) = sup

0<ε<N<∞

∣∣∣H (ε ,N)θ ( f )(x)

∣∣∣ .

We observe that for any fixed 0 < ε < N < ∞ and f ∈ Lp(Rn), H(ε ,N)θ ( f ) is well

defined almost everywhere. Indeed, by Minkowski’s integral inequality we obtain

∥∥H(ε ,N)θ ( f )

∥∥Lp(Rn)

≤ 2

π

∥∥ f∥∥

Lp(Rn)log

N

ε< ∞ ,

which implies that H(ε ,N)θ ( f )(x) is finite for almost all x ∈ Rn. Thus H

(∗∗)θ ( f ) is

well defined for f in⋃

1≤p<∞Lp(Rn).

Theorem 5.2.7. If Ω is odd and integrable over Sn−1, then TΩ and T(∗∗)Ω are Lp

bounded for all 1 < p < ∞. More precisely, TΩ initially defined on Schwartz func-

tions has a bounded extension on Lp(Rn) (which is also denoted by TΩ ).

Proof. Let e j be the usual unit vectors in Sn−1. The operator He1is obtained by

applying the Hilbert transform in the first variable followed by the identity operator

in the remaining variables. Clearly, He1is bounded on Lp(Rn) with norm equal to

that of the Hilbert transform on Lp(R). Next observe that the following identity is

valid for all matrices A ∈ O(n):

HA(e1)( f )(x) = He1( f A)(A−1x) . (5.2.18)

This implies that the Lp boundedness of Hθ can be reduced to that of He1. We

conclude that Hθ is Lp bounded for 1 < p < ∞ with norm bounded by the norm of

the Hilbert transform on Lp(R) for every θ ∈ Sn−1.

Identity (5.2.18) is also valid for H(ε ,N)θ and H

(∗∗)θ . Consequently, H

(∗∗)θ is

bounded on Lp(Rn) for 1 < p < ∞ with norm at most that of H(∗∗) on Lp(R).Next we realize a general singular integral TΩ with Ω odd as an average of the

directional Hilbert transforms Hθ . We start with f in⋃

1≤p<∞Lp(Rn) and the fol-

lowing identities:

∫

ε≤|y|≤N

Ω(y/|y|)|y|n f (x− y)dy = +

∫

Sn−1Ω(θ)

∫ N

r=εf (x− rθ)

dr

rdθ

= −∫

Sn−1Ω(θ)

∫ N

r=εf (x+ rθ)

dr

rdθ ,

where the first follows by switching to polar coordinates and the second one is a

consequence of the first one and the fact that Ω is odd via the change variables

θ → −θ . Averaging the two identities, we obtain

∫

ε≤|y|≤N

Ω(y/|y|)|y|n f (x− y)dy

=1

2

∫

Sn−1Ω(θ)

∫ N

r=ε

f (x− rθ)− f (x+ rθ)

rdr dθ

=π

2

∫

Sn−1Ω(θ)H

(ε ,N)θ ( f )(x)dθ .

(5.2.19)

It follows from the identity in (5.2.19) that

∫

ε≤|y|≤N

Ω(y/|y|)|y|n f (x− y)dy =

π

2

∫

Sn−1Ω(θ)H

(ε ,N)θ ( f )(x)dθ , (5.2.20)

from which we conclude that

T(∗∗)Ω ( f )(x)≤ π

2

∫

Sn−1|Ω(θ)|H (∗∗)

θ ( f )(x)dθ . (5.2.21)

Using the Lebesgue dominated convergence theorem, we see that for f in S (Rn),we can pass the limits as ε→ 0 and N→∞ inside the integral in (5.2.20), concluding

that

TΩ ( f )(x) =π

2

∫

Sn−1Ω(θ)Hθ ( f )(x)dθ , (5.2.22)

for f ∈ S (Rn). The Lp boundedness of TΩ and T(∗∗)Ω for Ω odd are then trivial

consequences of (5.2.22) and (5.2.21) via Minkowski’s integral inequality.

Corollary 5.2.8. The Riesz transforms R j and the maximal Riesz transforms R(∗)j

are bounded on Lp(Rn) for 1 < p < ∞.

Proof. The assertion follows from the fact that the Riesz transforms have odd ker-

nels. Since the kernel of R j decays like |x|−n near infinity, it follows that R(∗)j ( f ) is

well defined for f ∈ Lp(Rn). Since R(∗)j is pointwise bounded by 2R

(∗∗)j , the conclu-

sion follows from Theorem 5.2.7.

Remark 5.2.9. It follows from the proof of Theorem 5.2.7 and from Theorems 5.1.7

and 5.1.12 that whenever Ω is an odd function on Sn−1, we have

∥∥TΩ∥∥

Lp→Lp ≤∥∥Ω

∥∥L1

a p when p≥ 2,

a(p−1)−1 when 1 < p≤ 2,

∥∥T(∗∗)Ω

∥∥Lp→Lp ≤

∥∥Ω∥∥

L1

a p when p≥ 2,

a(p−1)−1 when 1 < p≤ 2,

for some a > 0 independent of p and the dimension.

5.2.4 Singular Integrals with Even Kernels

Since a general integrable function Ω on Sn−1 with mean value zero can be written

as a sum of an odd and an even function, it suffices to study singular integral opera-

tors TΩ with even kernels. For the rest of this section, fix an integrable even function

Ω on Sn−1 with mean value zero. The following idea is fundamental in the study of

such singular integrals. Proposition 5.1.16 implies that

TΩ =−n

∑j=1

R jR jTΩ . (5.2.23)

If R jTΩ were another singular integral operator of the form TΩ jfor some odd Ω j,

then the boundedness of TΩ would follow from that of TΩ jvia the identity (5.2.23)

and Theorem 5.2.7. It turns out that R jTΩ does have an odd kernel, but it may not be

integrable on Sn−1 unless Ω itself possesses an additional amount of integrability.

The amount of extra integrability needed is logarithmic, more precisely of this sort:

cΩ =∫

Sn−1|Ω(θ)| log+ |Ω(θ)|dθ < ∞ . (5.2.24)

Observe that ∥∥Ω∥∥

L1 ≤ cΩ + eωn−1 ≤Cn (cΩ +1) ,

which says that the norm ‖Ω‖L1 is always controlled by a dimensional constant

multiple of cΩ +1. The following theorem is the main result of this section.

Theorem 5.2.10. Let n ≥ 2 and let Ω be an even integrable function on Sn−1 with

mean value zero that satisfies (5.2.24). Then the corresponding singular integral

TΩ is bounded on Lp(Rn), 1 < p < ∞, with norm at most a dimensional constant

multiple of the quantity max((p−1)−2, p2

)(cΩ +1).

If the operator TΩ in Theorem 5.2.10 is weak type (1,1), then the estimate on the

Lp operator norm of TΩ can be improved to ‖TΩ‖Lp→Lp ≤ Cn(p− 1)−1 as p → 1.

This is indeed the case; see the historical comments at the end of this chapter.

Proof. Let WΩ be the distributional kernel of TΩ . We have that WΩ coincides with

the function Ω(x/|x|)|x|−n on Rn \0. Using Proposition 5.2.3 and the fact that Ωis an even function, we obtain the formula

WΩ (ξ ) =∫

Sn−1Ω(θ) log

1

|ξ ·θ | dθ , (5.2.25)

which implies that WΩ is itself an even function. Now, using Exercise 5.2.3 and

condition (5.2.24), we conclude that WΩ is a bounded function. Therefore, TΩ is L2

bounded. To obtain the Lp boundedness of TΩ , we use the idea mentioned earlier

involving the Riesz transforms. In view of (5.1.46), we have that

TΩ =−n

∑j=1

R jTj, (5.2.26)

where Tj = R jTΩ . Equality (5.2.26) makes sense as an operator identity on L2(Rn),since TΩ and each R j are well defined and bounded on L2(Rn).

The kernel of the operator Tj is the inverse Fourier transform of the distribu-

tion −iξ j

|ξ |WΩ (ξ ), which we denote by K j. At this point we know only that K j is

a tempered distribution whose Fourier transform is the function −iξ j

|ξ |WΩ (ξ ). Our

first goal is to show that K j coincides with an integrable function on an annulus. To

prove this assertion we write

WΩ =W 0Ω +W 1

Ω +W∞Ω ,

where W 0Ω is a distribution and W 1

Ω ,W∞Ω are functions defined by

⟨W 0Ω ,ϕ

⟩= lim

ε→0

∫

ε<|x|≤ 12

Ω(x/|x|)|x|n ϕ(x)dx ,

W 1Ω (x) =

Ω(x/|x|)|x|n χ 1

2≤|x|≤2,

W∞Ω (x) =

Ω(x/|x|)|x|n χ2<|x| .

We now fix a j ∈ 1,2, . . . ,n and we write

K j = K0j +K1

j +K∞j ,

where

K0j =

(− i

ξ j

|ξ |W0Ω (ξ )

)∨ ,

K1j =

(− i

ξ j

|ξ |W1Ω (ξ )

)∨ ,

K∞j =

(− i

ξ j

|ξ |W∞Ω (ξ )

)∨ .

Notice that K0j is well defined via Theorem 2.3.21.

Define the annulus

A = x ∈ Rn : 2/3 < |x|< 3/2.

For a smooth function φ supported in the annulus 2/3 < |x|< 3/2 we have

⟨K0

j ,φ⟩

=⟨(−i

ξ j

|ξ |W0Ω (ξ )

)∨,φ

⟩

=⟨− i

ξ j

|ξ |W0Ω (ξ ),φ

∨(ξ )⟩

=⟨W 0Ω (ξ ),−i

ξ j

|ξ |φ∨(ξ )

⟩

=⟨W 0Ω ,

(− i

ξ j

|ξ |φ∨(ξ )

)∧⟩

= −⟨W 0Ω , R j(φ)

⟩

= − limε→0

∫

ε<|y|<1/2

Ω(y/|y|)|y|n R j(φ)(−y)dy (Ω is even)

= −Γ (n+1

2)

πn+1

2

limε→0

∫

ε<|y|<1/2

Ω(y/|y|)|y|n

∫

Rn

y j− x j

|y− x|n+1φ(x)dxdy,

where the action of the distribution W 0Ω on R j(φ) is justified by fact that R j(φ)(y) is

smooth on the support of W 0Ω ; note |x− y| ≥ 1/6. Moreover, 〈W 0

Ω (ξ ),−iξ j

|ξ |φ∨(ξ )〉

should be interpreted as a convergent integral.

It follows that for x ∈ A, the absolute value of the convolution of W 0Ω with the

kernel of the Riesz transform R j is

∣∣∣∣∣Γ ( n+1

2)

πn+1

2

limε→0

∫

ε<|y|< 12

x j− y j

|x− y|n+1

Ω(y/|y|)|y|n dy

∣∣∣∣∣ (5.2.27)

=

∣∣∣∣∣Γ ( n+1

2)

πn+1

2

∫

|y|< 12

(x j− y j

|x− y|n+1− x j

|x|n+1

)Ω(y/|y|)|y|n dy

∣∣∣∣∣ (5.2.28)

≤∫

|y|≤ 12

Cn|y||Ω(y/|y|)||y|n dy

=C′n∥∥Ω

∥∥L1 ,

where we used the fact that Ω(y/|y|)|y|−n has integral zero over annuli of the form

ε < |y|< 12, the mean value theorem applied to the function x j|x|−(n+1), and the fact

that |x− y| ≥ 1/6 for x in the annulus A. We conclude that on A, K0j coincides with

the bounded function inside the absolute value in (5.2.27).

Likewise, for x ∈ A we have

Γ ( n+12)

πn+1

2

∣∣∣∣∫

|y|>2

x j− y j

|x− y|n+1

Ω(y/|y|)|y|n dy

∣∣∣∣ (5.2.29)

≤ Γ ( n+12)

πn+1

2

∫

|y|>2

1

|x− y|n|Ω(y/|y|)||y|n dy

≤ Γ ( n+12)

πn+1

2

∫

|y|>2

4n

|y|2n|Ω(y/|y|)|dy

=C∥∥Ω

∥∥L1 ,

from which it follows that on the annulus A, K∞j coincides with the bounded function

inside the absolute value in (5.2.29) or in (5.2.28).

Now observe that condition (5.2.24) gives that the function W 1Ω satisfies

∫

|x|≤2|W 1

Ω (x)| log+ |W 1Ω (x)|dx

≤∫ 2

1/2

∫

Sn−1

|Ω(θ)|rn

log+[2n|Ω(θ)|]dθrn−1 dr

r

≤ (log4)[n(log2)

∥∥Ω∥∥

L1 + cΩ]< ∞ .

Since the Riesz transform R j is countably subadditive and maps Lp to Lp with norm

at most 4(p−1)−1 for 1 < p < 2, it follows from Exercise 1.3.7 that K1j = R j(W

1Ω )

is integrable over the ball |x| ≤ 3/2 and moreover, it satisfies

∫

A|K1

j (x)|dx≤Cn

[∫

|x|≤2|W 1

Ω (x)| log+ |W 1Ω (x)|dx+1

]≤C′n(cΩ +1) .

Furthermore, since K j is homogeneous of degree zero, K j is a homogeneous

distribution of degree −n (Exercise 2.3.9). This means that for all test functions ϕand all λ > 0 we have ⟨

K j,δλ (ϕ)

⟩=

⟨K j,ϕ

⟩, (5.2.30)

where δ λ (ϕ)(x) = ϕ(λx). But for ϕ ∈C ∞0 supported in the annulus 3/4 < |x|< 4/3

and for λ in (8/9,9/8) we have that δ λ−1(ϕ) is supported in A and thus we can

express (5.2.30) as convergent integrals as follows:

∫

RnK j(x)ϕ(x)dx =

∫

RnK j(x)ϕ(λ

−1x)dx =

∫

Rnλ nK j(λx)ϕ(x)dx . (5.2.31)

From this it would be ideal to be able to directly obtain that K j(x) = λ nK j(λx) for all

8/9< |x|< 9/8 and 8/9< λ < 9/8, in particular when λ = |x|−1. But unfortunately,

we can only deduce that for every λ ∈ (8/9,9/8), K j(x) = λ nK j(λx) holds for all x

in the annulus except a set of measure zero that depends on λ . To be able to define

the restriction of K j on Sn−1, we employ a more delicate argument.

For any J subinterval of [8/9,9/8] we obtain from (5.2.31) that

∫

RnK j(x)ϕ(x)dx =

∫

Rn−∫

Jλ nK j(λx)dλ ϕ(x)dx ,

where integral with the slashed integral denotes the average of a function over the set

J. Since ϕ was an arbitrary C ∞0 function supported in the annulus 3/4 < |x|< 4/3,

it follows that for every J subinterval of [8/9,9/8], there is a null subset EJ of the

annulus A′ = x : 27/32 < |x|< 32/27 such that

K j(x) =−∫

Jλ nK j(λx)dλ (5.2.32)

for all x ∈ A′ \EJ .

Let J0 = [√

8/9,√

9/8]. We claim that there is a set of null subset E of A′ such

that for all x ∈ A′ \E we have

−∫

J0

λ nK j(λx)dλ =−∫

rJ0

λ nK j(λx)dλ (5.2.33)

for every r in J0. Indeed, let E be the union of ErJ0over all r in J0 ∩Q. Then in

view of (5.2.32), identity (5.2.33) holds for x ∈ A′ \E and J0∩Q. But for a fixed x in

A′ \E, the function of r on the right hand side of (5.2.33) is constant on the rationals

and is also continuous (in r), hence it must be constant for all r ∈ J0. Thus the claim

follows since both sides of (5.2.33) are equal to (5.2.32).

Writing x = δθ , where 27/32 < δ < 32/27 and θ ∈ Sn−1, it follows by Fubini’s

theorem that there is a δ ∈ (27/32,32/27) (in fact almost all δ have this property)

such that

−∫

J0

λ nK j(λδθ)dλ =−∫

rJ0

λ nK j(λδθ)dλ (5.2.34)

for almost all θ ∈ Sn−1 and all r ∈ J0. We fix such a δ , which we denote δ0.

We now define a function Ω j on Sn−1 by setting

Ω j(θ) =−∫

J0

δ n0 λ

nK j(λδ0θ)dλ =−∫

rJ0

δ n0 λ

nK j(λδ0θ)dλ

for all r ∈ J0. The function Ω j is defined almost everywhere and is integrable over

Sn−1, since K j is integrable over the annulus A.

Let e1 = (1,0, . . . ,0). LetΨ be a C ∞0 (Rn) nonzero, nonnegative, radial, and sup-

ported in the annulus 32/(27√

2)< |x|< 27√

2/32 around Sn−1. We start with

Ω j(θ) =−∫

r−1J0

δ n0 λ

nK j(λδ0θ)dλ =−∫

J0

δ n0 rnλ nK j(rλδ0θ)dλ ,

which holds for all r ∈ J0, we multiply by Ψ(re1), and we integrate over Sn−1 and

over (0,∞) with respect to the measure dr/r. We obtain

∫ ∞

0Ψ(re1)

dr

r

∫

Sn−1Ω j(θ)dθ = −

∫

J0

∫ ∞

0

∫

Sn−1δ n

0 λnK j(λδ0rθ)Ψ(re1)r

ndθdr

rdλ

= −∫

J0

∫

Rnδ n

0 λnK j(λδ0x)Ψ(x)dxdλ

= −∫

J0

∫

RnK j(x)Ψ((λδ0)

−1x)dxdλ

= −∫

J0

⟨K j,Ψ

⟩dλ ,

=⟨K j,Ψ

⟩

in view of the homogeneity of K j. But also, for some constant c′Ψ we have

⟨K j,Ψ

⟩=

⟨K j,Ψ

⟩=

∫

Rn

−iξ j

|ξ | WΩ (ξ )Ψ(ξ )dξ = c′Ψ

∫

Sn−1

−iθ j

|θ | WΩ (θ)dθ = 0,

since by (5.2.25),−iξ j

|ξ | WΩ (ξ ) is an odd function. We conclude that Ω j has mean

value zero over Sn−1.

Thus Ω ∈ L1(Sn−1) has mean value zero and the distribution WΩ jis well defined.

We claim that

K j =WΩ j. (5.2.35)

To establish (5.2.35), we show first that⟨K j,ϕ

⟩=

⟨WΩ j

,ϕ⟩

whenever ϕ is sup-

ported in the annulus 8/9 < |x|< 9/8. Using (5.2.32) we have

∫

RnK j(x)ϕ(x)dx =

∫

Rn−∫

J0

K j(δ0λx)δ n0 λ

ndλ ϕ(x)dx

=

∫ ∞

0

∫

Sn−1−∫

J0

K j(δ0λ rθ)δ n0 λ

nrn dλ ϕ(rθ)dθdr

r

=∫ ∞

0

∫

Sn−1−∫

rJ0

K j(δ0λ′θ)δ n

0 (λ′)ndλ ′ϕ(rθ)dθ

dr

r

=∫ ∞

0

∫

Sn−1Ω j(θ)ϕ(rθ)dθ

dr

r

=⟨WΩ j

,ϕ⟩,

having used (5.2.34) in the second to last equality.

Given a general C ∞0 function ϕ whose support is contained in an annulus of the

form M−1 < |x|<M, for some M > 0, via a smooth partition of unity, we write ϕ as a

finite sum of smooth functions ϕk whose supports are contained in annuli of the form

8s/9 < |x| < 9s/8 for some s > 0. These annuli can be brought inside the annulus

8/9 < |x|< 9/8 by a dilation. Since both K j and WΩ jare homogeneous distributions

of degree −n and agree on the annulus 8/9 < |x| < 9/8 they must agree on annuli

8s/9 < |x| < 9s/8. Consequently, 〈K j,ϕ〉 = 〈WΩ j,ϕ〉 for all ϕ ∈ C ∞

0 (Rn \ 0).Therefore, K j−WΩ j

is supported at the origin, and since it is homogeneous of degree

−n, it must be equal to bδ0, a constant multiple of the Dirac mass. But K j is an

odd function and hence K j is also odd. It follows that WΩ jis an odd function on

Rn \0, which implies that Ω j is an odd function. We say that u ∈S ′(Rn) is odd

if u = −u, where u is defined by 〈u,ψ〉 = 〈u, ψ〉 for all ψ ∈ S (Rn) and ψ(x) =ψ(−x). We have that K j−WΩ j

is an odd distribution, and thus bδ0 must be an odd

distribution. But if bδ0 is odd, then b= 0. We conclude that for each j there exists an

odd integrable function Ω j on Sn−1 with ‖Ω j‖L1 controlled by a constant multiple

of cΩ +1 such that (5.2.35) holds.

Then we use (5.2.26) and (5.2.35) to write

TΩ =−n

∑j=1

R jTΩ j,

and appealing to the boundedness of each TΩ j(Theorem 5.2.7) and to that of the

Riesz transforms, we obtain the required Lp boundedness for TΩ .

We note that Theorem 5.2.10 holds for all Ω ∈ L1(Sn−1) that satisfy (5.2.24), not

necessarily even Ω . Simply write Ω = Ωe +Ωo , where Ωe is even and Ωo is odd,

and check that condition (5.2.24) holds for Ωe.

5.2.5 Maximal Singular Integrals with Even Kernels

We have the corresponding theorem for maximal singular integrals.

Theorem 5.2.11. Let Ω be an even integrable function on Sn−1 with mean value

zero that satisfies (5.2.24). Then the corresponding maximal singular integral T(∗∗)Ω ,

defined in (5.2.4), is bounded on Lp(Rn) for 1 0

1

2a

∫

|r|≤a| f (x− rθ)|dr . (5.2.36)

In view of Exercise 5.2.5 we have that Mθ is bounded on Lp(Rn) with norm at most

3 p(p−1)−1.

Fix Φ a smooth radial function such that Φ(x) = 0 for |x| ≤ 1/4, Φ(x) = 1 for

|x| ≥ 3/4, and 0≤Φ(x)≤ 1 for all x in Rn. For f ∈ Lp(Rn) and 0 < ε < N < ∞ we

introduce the smoothly truncated singular integral

T(ε ,N)Ω ( f )(x) =

∫

Rn

Ω(

y|y|)

|y|n(Φ

(yε

)−Φ

(yN

))f (x− y)dy

and the corresponding maximal singular integral operator

T(∗∗)Ω ( f ) = sup

0<N<∞sup

0<ε<N

|T (ε ,N)Ω ( f )| . (5.2.37)

Computing the supremum in (5.2.37), we first consider the case where N > 4ε .

For f in Lp(Rn) (for some 1 < p < ∞), we have

∣∣T (ε ,N)Ω ( f )(x)−T

(ε ,N)Ω ( f )(x)

∣∣

=

∣∣∣∣∫

ε4≤|y|≤ε

Ω(

y|y|)

|y|n Φ(

yε

)f (x−y)dy−

∫

N4 ≤|y|≤N

Ω(

y|y|)

|y|n Φ(

yN

)f (x−y)dy

∣∣∣∣

≤[ ∫

ε4≤|y|≤ε

|Ω(

y|y|)|

|y|n | f (x− y)|dy+∫

N4 ≤|y|≤N

|Ω(

y|y|)|

|y|n | f (x− y)|dy

]

≤∫

Sn−1

|Ω(θ)|[

4

ε

∫ ε

ε4

| f (x− rθ)|dr+4

N

∫ N

N4

| f (x− rθ)|dr

]dθ

≤ 16

∫

Sn−1|Ω(θ)|Mθ ( f )(x)dθ .

Now if N ≤ 4ε , then the function Φ(

yε

)−Φ

(yN

)−χε≤|y|≤N is bounded by 3 and is

supported in the annulus ε4≤ |y| ≤ 4ε . In this case we obtain

∣∣T (ε ,N)Ω ( f )(x)−T

(ε ,N)Ω ( f )(x)

∣∣ ≤ 3

∫

Sn−1

|Ω(θ)|∫ 4ε

ε4

| f (x− rθ)| dr

rdθ

≤ 96

∫


We deduce from these estimates that

sup0<ε<N<∞

∣∣T (ε ,N)Ω ( f )(x)−T

(ε ,N)Ω ( f )(x)

∣∣≤ 96

∫


Using the result of Exercise 5.2.5 we conclude that

∥∥T(∗∗)Ω ( f )−T

(∗∗)Ω ( f )

∥∥Lp ≤ 600

∥∥Ω∥∥

L1 max(p,(p−1)−1)∥∥ f

∥∥Lp .

This implies that it suffices to obtain the required Lp bound for the smoothly trun-

cated maximal singular integral operator T(∗∗)Ω .

Let K j,Ω j, and Tj be as in the previous theorem, and let Fj be the Riesz transform

of the functionΩ(x/|x|)Φ(x)|x|−n. Let f ∈ Lp(Rn). A calculation yields the identity

T(ε ,N)Ω ( f )(x) =

∫

Rn

[1

εn

Ω( yε /|

yε |)

| yε |n

Φ( yε )−

1

Nn

Ω( yN/| y

N|)

| yN|n Φ( y

N)

]f (x− y)dy

= −( n

∑j=1

[1εn Fj

( ·ε

)− 1

Nn Fj

( ·N

)]∗R j( f )

)(x) ,

where in the last step we used Proposition 5.1.16. Therefore we may write

−T(ε ,N)Ω ( f )(x) =

n

∑j=1

∫

Rn

[1εn Fj

(x−yε

)− 1

Nn Fj

(x−yN

)]R j( f )(y)dy

= A(ε ,N)1 ( f )(x)+A

(ε ,N)2 ( f )(x)+A

(ε ,N)3 ( f )(x) ,

(5.2.38)

where

A(ε ,N)1 ( f )(x) =

n

∑j=1

1

εn

∫

|x−y|≤εFj

(x−yε

)R j( f )(y)dy

−n

∑j=1

1

Nn

∫

|x−y|≤NFj

(x−yN

)R j( f )(y)dy ,

A(ε ,N)2 ( f )(x) =

n

∑j=1

∫

Rn

[1εn χ|x−y|>ε

Fj

(x−yε

)−K j

(x−yε

)

− 1Nn χ|x−y|>N

Fj

(x−yN

)−K j

(x−yN

)]R j( f )(y)dy ,

A(ε ,N)3 ( f )(x) =

n

∑j=1

∫

Rn

[1εn χ|x−y|>εK j

(x−yε

)− 1

Nn χ|x−y|>NK j

(x−yN

)]R j( f )(y)dy .

It follows from the definitions of Fj and K j that

Fj(z)−K j(z) =Γ ( n+1

2)

πn+1

2

limε→0

∫

ε≤|y|

Ω(y/|y|)|y|n

(Φ(y)−1

) z j− y j

|z− y|n+1dy

=Γ ( n+1

2)

πn+1

2

∫

|y|≤ 34

Ω(y/|y|)|y|n

(Φ(y)−1

) z j− y j

|z− y|n+1− z j

|z|n+1

dy

whenever |z| ≥ 1. But using the mean value theorem, the last expression is easily

seen to be bounded by

Cn

∫

|y|≤ 34

Ω(y/|y|)|y|n

|y||z|n+1

dy =C′n∥∥Ω

∥∥L1 |z|−(n+1) ,

whenever |z| ≥ 1. Using this estimate, we obtain that the jth term in A(ε ,N)2 ( f )(x) is

bounded by

Cn

‖Ω‖L1

εn

∫

|x−y|>ε

|R j( f )(y)|dy

(|x− y|/ε)n+1≤Cn

2‖Ω‖L1

2−nεn

∫

Rn

|R j( f )(y)|dy(1+ |x−y|

ε

)n+1.

It follows that for functions f in Lp we have

sup0<ε<N<∞

|A(ε ,N)2 ( f )| ≤Cn

∥∥Ω∥∥

L1 M(R j( f )) ,

in view of Theorem 2.1.10. (M here is the Hardy–Littlewood maximal operator.)

By Theorem 2.1.6, M maps Lp(Rn) to itself with norm bounded by a dimensional

constant multiple of max(1,(p−1)−1). Since by Remark 5.2.9 the norm∥∥R j

∥∥Lp→Lp

is controlled by a dimensional constant multiple of max(p,(p−1)−1), it follows that

∥∥ sup0<ε<N<∞

|A(ε ,N)2 ( f )|

∥∥Lp ≤Cn

∥∥Ω∥∥

L1 max(p,(p−1)−1)∥∥ f

∥∥Lp . (5.2.39)

Next, recall that in the proof of Theorem 5.2.10 we showed that

K j(x) =Ω j(x/|x|)|x|n ,

where Ω j are integrable functions on Sn−1 that satisfy

∥∥Ω j

∥∥L1 ≤Cn(cΩ +1) . (5.2.40)

Consequently, for functions f in Lp(Rn) we have

sup0<ε<N<∞

|A(ε ,N)3 ( f )| ≤ 2

n

∑j=1

T(∗∗)Ω j

(R j( f )) ,

and by Remark 5.2.9 this last expression has Lp norm at most a dimensional constant

multiple of∥∥Ω j

∥∥L1 max(p,(p−1)−1)

∥∥R j( f )∥∥

Lp . It follows that

∥∥∥ sup0<ε<N<∞

|A(ε ,N)3 ( f )|

∥∥∥Lp≤Cn max(p2,(p−1)−2)(cΩ +1)

∥∥ f∥∥

Lp . (5.2.41)

Finally, we turn our attention to the term A(ε ,N)1 ( f ). To prove the required esti-

mate, we first show that there exist nonnegative homogeneous of degree zero func-

tions G j on Rn that satisfy

|Fj(x)| ≤ G j(x) when |x| ≤ 1 (5.2.42)

and ∫

Sn−1|G j(θ)|dθ ≤Cn(cΩ +1) . (5.2.43)

To prove (5.2.42), first note that if |x| ≤ 1/8, then

|Fj(x)| =Γ ( n+1

2)

πn+1

2

∣∣∣∣∫

Rn

Ω(y/|y|)|y|n Φ(y)

x j− y j

|x− y|n+1dy

∣∣∣∣

≤Cn

∫

|y|≥ 14

|Ω(y/|y|)||y|2n

dy

≤C′n∥∥Ω

∥∥L1 .


We now fix an x satisfying 1/8≤ |x| ≤ 1 and we write

|Fj(x)| ≤Φ(x)|K j(x)|+ |Fj(x)−Φ(x)K j(x)|

≤ |K j(x)|+Γ ( n+1

2)

πn+1

2

∣∣∣∣ limε→0

∫

|y|>ε

x j− y j

|x− y|n+1

(Φ(y)−Φ(x)

)Ω(y/|y|)|y|n dy

∣∣∣∣

= |K j(x)|+Γ ( n+1

2)

πn+1

2

(P1(x)+P2(x)+P3(x)

),

where

P1(x) =

∣∣∣∣∫

|y|≤ 116

(x j− y j

|x− y|n+1− x j

|x|n+1

)(Φ(y)−Φ(x)

)Ω(y/|y|)|y|n dy

∣∣∣∣ ,

P2(x) =

∣∣∣∣∫

116≤|y|≤2

x j− y j

|x− y|n+1

(Φ(y)−Φ(x)

)Ω(y/|y|)|y|n dy

∣∣∣∣ ,

P3(x) =

∣∣∣∣∫

|y|≥2

x j− y j

|x− y|n+1

(Φ(y)−Φ(x)

)Ω(y/|y|)|y|n dy

∣∣∣∣ .

But since 1/8≤ |x| ≤ 1, we see that

P1(x)≤Cn

∫

|y|≤ 116

|y||x|n+1

|Ω(y/|y|)||y|n dy≤C′n

∥∥Ω∥∥

L1

and that

P3(x)≤Cn

∫

|y|≥2

|Ω(y/|y|)||y|2n

dy≤C′n∥∥Ω

∥∥L1 .

For P2(x) we use the estimate |Φ(y)−Φ(x)| ≤C|x− y| to obtain

P2(x) ≤∫

116≤|y|≤2

C

|x− y|n−1

|Ω(y/|y|)||y|n dy

≤ 4C

∫

116≤|y|≤2

|Ω(y/|y|)||x− y|n−1|y|n− 1

2

dy

≤ 4C

∫

Rn

|Ω(y/|y|)||x− y|n−1|y|n− 1

2

dy .

Recall that K j(x) =Ω j(x/|x|)|x|−n. We now set

G j(x) =Cn

(∥∥Ω∥∥

L1 +∣∣∣Ω j

( x

|x|)∣∣∣+ |x|n− 3

2

∫

Rn

|Ω(y/|y|)|dy

|x− y|n−1|y|n− 12

)(5.2.44)

and we observe that G j is a homogeneous of degree zero function, it satisfies

(5.2.42), and it is integrable over the annulus 12≤ |x| ≤ 2. To verify the last assertion,

we split up the double integral

I =∫

12≤|x|≤2

∫

Rn

|Ω(y/|y|)|dy

|x− y|n−1|y|n− 12

dx

into the pieces 1/4≤ |y| ≤ 4, |y|> 4, and |y|< 1/4. The part of I where 1/4≤ |y| ≤ 4

is pointwise bounded by a constant multiple of

∫

14≤|y|≤4

∣∣∣Ω( y

|y|)∣∣∣

∫

12≤|x|≤2

dx

|y− x|n−1dy≤

∫

14≤|y|≤4

∣∣∣Ω( y

|y|)∣∣∣

∫

|x−y|≤6

dx

|y− x|n−1dy ,

which is pointwise controlled by a constant multiple of ‖Ω‖L1 . In the part of I where

|y| > 4 we use that |x− y|−n+1 ≤ (|y|/2)−n+1 to obtain rapid decay in y and hence

a bound by a constant multiple of ‖Ω‖L1 . Finally, in the part of I where |y| < 1/4

we use that |x− y|−n+1 ≤ (1/4)−n+1, and then we also obtain a similar bound. It

follows from (5.2.44) and (5.2.40) that

∫

12≤|x|≤2

|G j(x)|dx≤Cn

(‖Ω‖L1 +‖Ω j‖L1 +‖Ω‖L1

)≤Cn(cΩ +1).

Since G j is homogeneous of degree zero, we deduce (5.2.43).

To complete the proof, we argue as follows:

sup0<ε<N<∞

|A(ε ,N)1 ( f )(x)|

≤ 2supε>0

n

∑j=1

1

εn

∫

|z|≤ε|Fj(z)| |R j( f )(x− z)|dz

≤ 2supε>0

n

∑j=1

1

εn

∫ ε

r=0

∫

Sn−1|Fj(rθ)| |R j( f )(x− rθ)|rn−1 dθ dr

≤ 2n

∑j=1

∫

Sn−1|G j(θ)|

supε>0

1

εn

∫ ε

r=0|R j( f )(x− rθ)|rn−1 dr

dθ

≤ 4n

∑j=1

∫

Sn−1|G j(θ)|Mθ (R j( f ))(x)dθ .

Using (5.2.43) together with the Lp boundedness of the Riesz transforms and of Mθ

we obtain∥∥∥ sup

0<ε<N<∞|A(ε ,N)

1 ( f )|∥∥∥

Lp≤Cn max(p,(p−1)−2)(cΩ +1)

∥∥ f∥∥

Lp . (5.2.45)

Combining (5.2.45), (5.2.39), and (5.2.41), we obtain the required conclusion.

The following corollary is a consequence of Theorem 5.2.11.

Corollary 5.2.12. Let Ω be as in Theorem 5.2.11. Then for 1 < p < ∞ and f in

Lp(Rn) the functions T(ε ,N)Ω ( f ) converge to TΩ ( f ) in Lp and almost everywhere as

ε → 0 and N → ∞.

Proof. The a.e. convergence is a consequence of Theorem 2.1.14. The Lp conver-

gence is a consequence of the Lebesgue dominated convergence theorem since for

f ∈ Lp(Rn) we have that |T (ε ,N)Ω ( f )| ≤ T

(∗∗)Ω ( f ) and T

(∗∗)Ω ( f ) is in Lp(Rn).

Exercises

5.2.1. Show that the directional Hilbert transform Hθ is given by convolution with

the distribution wθ in S ′(Rn) defined by

⟨wθ ,ϕ

⟩=

1

πp.v.

∫ +∞

−∞

ϕ(tθ)

tdt.

Compute the Fourier transform of wθ and prove that Hθ maps L1(Rn) to L1,∞(Rn).

5.2.2. Extend the definitions of WΩ and TΩ toΩ = dµ a finite signed Borel measure

on Sn−1 with mean value zero. Compute the Fourier transform of Wdµ and find a

necessary and sufficient condition on measures dµ so that Tdµ is L2 bounded. Notice

that the directional Hilbert transform Hθ is a special case of such an operator Tdµ .

5.2.3. Use the inequality AB ≤ A logA+ eB for A ≥ 1 and B > 0 to prove that if

Ω satisfies (5.2.24) then it must satisfy (5.2.16). Conclude that if |Ω | log+ |Ω | is in

L1(Sn−1), then TΩ is L2 bounded.[Hint: Use that

∫Sn−1 |ξ ·θ |−α dθ converges when α < 1. See Appendix D.3.

]

5.2.4. Let Ω be a nonzero integrable function on Sn−1 with mean value zero. Let

f ≥ 0 be nonzero and integrable over Rn. Prove that TΩ ( f ) is not in L1(Rn).[Hint: Show that TΩ ( f ) cannot be continuous at zero.

]

5.2.5. Let θ ∈ Sn−1. Use an identity similar to (5.2.18) to show that the maximal

operators

supa>0

1

a

∫ a

0| f (x− rθ)|dr , sup

a>0

1

2a

∫ +a

−a| f (x− rθ)|dr

are Lp(Rn) bounded for 1 0

1

vnRn

∫

|y|≤R|Ω(y/|y|)| | f (x− y)|dy .

Apply the method of rotations to prove that MΩ maps Lp(Rn) to itself for 1 < p<∞.

5.2.7. Let Ω(x,θ) be a function on Rn×Sn−1 satisfying

(a) Ω(x,−θ) =−Ω(x,θ) for all x and θ .

(b) supx |Ω(x,θ)| is in L1(Sn−1).

Use the method of rotations to prove that

TΩ ( f )(x) = p.v.

∫

Rn

Ω(x,y/|y|)|y|n f (x− y)dy

is bounded on Lp(Rn) for 1 0 such that for every

complex-valued C 2(R2) function f with compact support we have the bound

∥∥∂x1f∥∥

Lp +∥∥∂x2

f∥∥

Lp ≤ Ap

∥∥∂x1f + i∂x2

f∥∥

Lp .

5.2.10. (a) Let ∆ = ∑nj=1 ∂

2x j

be the usual Laplacian on Rn. Prove that for all 1 0 such that for all C 2 functions f with compact

support we have the bound

∥∥∂x j∂xk

f∥∥

Lp ≤ Ap

∥∥∆ f∥∥

Lp .

(b) Let ∆m =

m times︷︸︸︷∆ · · · ∆ . Show that for any 1 0 such

that for all f of class C 2m with compact support and all differential monomials ∂αxof order |α|= 2m we have

∥∥∂αx f∥∥

Lp ≤Cp

∥∥∆m f∥∥

Lp .

5.2.11. Use the same idea as in Lemma 5.2.5 to show that if f is continuous on

[0,∞), differentiable in (0,∞), and satisfies

limN→∞

∫ Na

N

f (u)

udu = 0

for all a > 0, then

limε→0N→∞

∫ N

ε

f (at)− f (t)

tdt = f (0) log

1

a.

5.2.12. Let Ωo be an odd integrable function on Sn−1 and Ωe an even function on

Sn−1 that satisfies (5.2.24). Let f be a function supported in a ball B in Rn. Prove

that

(a) If | f | log+ | f | is integrable over a ball B, then TΩo( f ) and T

(∗∗)Ωo

( f ) are integrable

over B.

(b) If | f |(log+ | f |)2 is integrable over a ball B, then TΩe( f ) and T

(∗∗)Ωe

( f ) are inte-

grable over B.[Hint: Use Exercise 1.3.7.

]

5.3 Calderon–Zygmund Decomposition and Singular Integrals 355

5.2.13. ([324]) Let Ω be integrable on Sn−1 with mean value zero. Use Jensen’s

inequality to show that for some C > 0 and every radial function f ∈ L2(Rn) we

have ∥∥TΩ ( f )∥∥

L2 ≤C∥∥ f

∥∥L2 .

This inequality subsumes that TΩ is well defined on radial L2(Rn) functions.

5.3 The Calderon–Zygmund Decomposition and Singular

Integrals

The behavior of singular integral operators on L1(Rn) is a more subtle issue than

that on Lp for 1 < p < ∞. It turns out that singular integrals are not bounded from

L1 to L1. See Example 5.1.3 and also Exercise 5.2.4. In this section we see that

singular integrals map L1 into the larger space L1,∞. This result strengthens their Lp

boundedness.

5.3.1 The Calderon–Zygmund Decomposition

To make some advances in the theory of singular integrals, we need to introduce

the Calderon–Zygmund decomposition. This is a powerful stopping-time construc-

tion that has many other interesting applications. We have already encountered an

example of a stopping-time argument in Section 2.1.

Recall that a dyadic cube in Rn is the set

[2km1,2k(m1 +1))×·· ·× [2kmn,2

k(mn +1)) ,

where k,m1, . . . ,mn ∈Z. Two dyadic cubes are either disjoint or related by inclusion.

Theorem 5.3.1. Let f ∈ L1(Rn) and α > 0. Then there exist functions g and b on

Rn such that

(1) f = g+b.

(2) ‖g‖L1 ≤ ‖ f‖L1 and ‖g‖L∞ ≤ 2nα .

(3) b = ∑ j b j, where each b j is supported in a dyadic cube Q j. Furthermore, the

cubes Qk and Q j are disjoint when j = k.

(4)

∫

Q j

b j(x)dx = 0.

(5) ‖b j‖L1 ≤ 2n+1α|Q j|.

(6) ∑ j |Q j| ≤ α−1‖ f‖L1 .


Remark 5.3.2. This decomposition is called the Calderon–Zygmund decomposition

of f at height α . The function g is called the good function of the decomposition,

since it is both integrable and bounded; hence the letter g. The function b is called

the bad function, since it contains the singular part of f (hence the letter b), but it

is carefully chosen to have mean value zero. It follows from (1) and (2) that the bad

function b is integrable and satisfies

∥∥b∥∥

L1 ≤∥∥ f

∥∥L1 +

∥∥g∥∥

L1 ≤ 2∥∥ f

∥∥L1 .

By (2) the good function is integrable and bounded; hence it lies in all the Lp spaces

for 1≤ p≤ ∞. More specifically, we have the following estimate:

∥∥g∥∥

Lp ≤∥∥g

∥∥ 1p

L1

∥∥g∥∥1− 1

p

L∞ ≤∥∥ f

∥∥ 1p

L1(2nα)1− 1

p = 2np′ α

1p′∥∥ f

∥∥ 1p

L1 . (5.3.1)

Proof. Decompose Rn into a mesh of disjoint dyadic cubes of the same size such

that

|Q| ≥ 1

α

∥∥ f∥∥

L1

for every cube Q in the mesh. Call these cubes of zero generation. Subdivide each

cube of zero generation into 2n congruent cubes by bisecting each of its sides. We

now have a new mesh of dyadic cubes, which we call of generation one. Select a

cube Q of generation one if

1

|Q|

∫

Q| f (x)|dx > α. (5.3.2)

Let S(1) be the set of all selected cubes of generation one. Now subdivide each

nonselected cube of generation one into 2n congruent subcubes by bisecting each

side and call these cubes of generation two. Then select all cubes Q of generation

two if (5.3.2) holds. Let S(2) be the set of all selected cubes of generation two. Repeat

this procedure indefinitely.

The set of all selected cubes⋃∞

m=1 S(m) is countable and is exactly the set of the

cubes Q j proclaimed in the proposition. Note that in some instances this set may

be empty, in which case b = 0 and g = f . Let us observe that the selected cubes

are disjoint, for otherwise some Qk would be a proper subset of some Q j, which is

impossible since the selected cube Q j was never subdivided. Now define

b j =

(f − 1

|Q j|

∫

Q j

f dx

)χQ j

,

b = ∑ j b j, and g = f −b.

For a selected cube Q j there exists a unique nonselected cube Q′ with twice its

side length that contains Q j. Let us call this cube the parent of Q j. Since the parent

Q′ of Q j was not selected, we have |Q′|−1∫

Q′ | f |dx≤ α . Then


1

|Q j|

∫

Q j

| f (x)|dx≤ 1

|Q j|

∫

Q′| f (x)|dx =

2n

|Q′|

∫

Q′| f (x)|dx≤ 2nα.

Consequently,

∫

Q j

|b j|dx≤∫

Q j

| f |dx+ |Q j|∣∣∣∣

1

|Q j|

∫

Q j

f dx

∣∣∣∣≤ 2

∫

Q j

| f |dx≤ 2n+1α|Q j| ,

which proves (5). To prove (6), simply observe that

∑j

|Q j| ≤1

α ∑j

∫

Q j

| f |dx =1

α

∫⋃

j Q j

| f |dx≤ 1

α

∥∥ f∥∥

L1 .

Next we need to obtain the estimates concerning g. We obviously have

g =

⎧⎨⎩

f on Rn \⋃ j Q j,

1|Q j |

∫

Q j

f dx on Q j.(5.3.3)

On the cube Q j, g is equal to the constant |Q j|−1∫

Q jf dx, and this is bounded by

2nα . It suffices to show that g is bounded outside the union of the Q j’s. Indeed, for

each x ∈ Rn \⋃ j Q j and for each k = 0,1,2, . . . there exists a unique nonselected

dyadic cube Q(k)x of generation k that contains x. Then for each k ≥ 0, we have

∣∣∣∣∣1

|Q(k)x |

∫

Q(k)x

f (y)dy

∣∣∣∣∣≤1

|Q(k)x |

∫

Q(k)x

| f (y)|dy≤ α.

The intersection of the closures of the cubes Q(k)x is the singleton x. Using Corol-

lary 2.1.16, we deduce that for almost all x ∈ Rn \⋃ j Q j we have

f (x) = limk→∞

1

|Q(k)x |

∫

Q(k)x

f (y)dy .

Since these averages are at most α , we conclude that | f | ≤ α a.e. on Rn \⋃ j Q j,

hence |g| ≤ α a.e. on this set. Finally, it follows from (5.3.3) that ‖g‖L1 ≤ ‖ f‖L1 .

This finishes the proof of the theorem.

We now apply the Calderon–Zygmund decomposition to obtain weak type (1,1)bounds for a wide class of singular integral operators that includes the operators TΩwe studied in the previous section.

5.3.2 General Singular Integrals

The kernels of the general singular integrals we will study are tempered distributions

that coincide with functions away from the origin. The setup as follows. Let K be

a measurable function defined on Rn \0 that is integrable on compact subsets of

Rn \0 and satisfies the size condition

supR>0

∫

R≤|x|≤2R|K(x)|dx = A1 < ∞ . (5.3.4)

This condition is less restrictive than the standard size estimate

supx∈Rn

|x|n|K(x)|< ∞ , (5.3.5)

but it is strong enough to capture size properties of kernels K(x) = Ω(x/|x|)/|x|n,

where Ω ∈ L1(Sn−1). We also note that condition (5.3.4) is equivalent to

supR>0

1

R

∫

|x|≤R|K(x)| |x|dx < ∞ . (5.3.6)

See Exercise 5.3.1.

The size condition (5.3.4) is sufficient to make the restriction of K(x) on |x|> δa tempered distribution (for any δ > 0). Indeed, for ϕ ∈S (Rn) we have

∫

|x|≥1|K(x)ϕ(x)|dx ≤

∞

∑m=0

∫

2m+1≥|x|≥2m

|K(x)|(1+ |x|)N |ϕ(x)|(1+2m)N

dx

≤∞

∑m=0

A1

(1+2m)Nsupx∈Rn

(1+ |x|)N |ϕ(x)| ,

and this expression is bounded by a constant times a finite sum of Schwartz semi-

norms of ϕ .

We are interested in tempered distributions W on Rn that extend the function K

defined on Rn \0 and have the form

〈W,ϕ〉= limj→∞

∫

|x|≥δ j

K(x)ϕ(x)dx, ϕ ∈S (Rn), (5.3.7)

for some sequence δ j ↓ 0 as j→ ∞. It is not hard to see that there exists a tempered

distribution W satisfying (5.3.7) for all ϕ ∈S (Rn) if and only if

limj→∞

∫

1≥|x|≥δ j

K(x)dx = L (5.3.8)

exists. See Exercise 5.3.2. If such a distribution W exists it may not be unique, since

it depends on the choice of the sequence δ j. Two different sequences tending to zero

may give two different tempered distributions W of the form (5.3.7), both coinciding

with the function K on Rn \0. See Example 5.4.2 and Remark 5.4.3.

If condition (5.3.8) is satisfied, we can define


∫

j≥|x|≥δ j

K(x)ϕ(x)dx (5.3.9)

and the limit exists as j→ ∞ for all ϕ ∈S (Rn) and is equal to

〈W,ϕ〉=∫

|x|≤1K(x)(ϕ(x)−ϕ(0))dx+ϕ(0)L+

∫

|x|≥1K(x)ϕ(x)dx .

Moreover, the previous calculations show that W is an element of S ′(Rn).Next we assume that the given function K on Rn \0 satisfies a certain smooth-

ness condition. There are three kinds of smoothness conditions that we encounter:

first, the gradient condition

|∇K(x)| ≤ A2|x|−n−1, x = 0; (5.3.10)

next, the weaker Lipschitz condition,

|K(x− y)−K(x)| ≤ A2|y|δ|x|n+δ , whenever |x| ≥ 2|y|; (5.3.11)

and finally the even weaker smoothness condition

supy =0

∫

|x|≥2|y||K(x− y)−K(x)|dx = A2 , (5.3.12)

for some A2 <∞. One should verify that (5.3.12) is a weaker condition than (5.3.11),

which in turn is weaker than (5.3.10). Condition (5.3.12) is often referred to as

Hormander’s condition.

5.3.3 Lr Boundedness Implies Weak Type (1,1) Boundedness

This next theorem provides the most classical application of the Calderon–Zygmund

decomposition.

Theorem 5.3.3. Let K be a function on Rn \0 that satisfies (5.3.4)1 and (5.3.12)

for some A1,A2 < ∞. Let W be an element of S ′(Rn) related to K as in (5.3.7).

Suppose that the operator T given by convolution with W has a bounded extension

that maps Lr(Rn) to itself with norm B for some 1 < r≤∞. Then T has an extension

that maps L1(Rn) to L1,∞(Rn) with norm

1 this condition could be replaced by the assumption that K is integrable over any compact set that

does not contain the origin.

∥∥T∥∥

L1→L1,∞ ≤Cn (A2 +B), (5.3.13)

and T also extends to a bounded operator from Lp(Rn) to itself for 1 < p < ∞ with

norm ∥∥T∥∥

Lp→Lp ≤C′n max(

p,(p−1)−1)(A2 +B), (5.3.14)

where Cn,C′n are constants that depend on the dimension but not on r or p.

Proof. We discuss the case r < ∞ and we refer to Exercise 5.3.7 for the case r = ∞.

Let α > 0 be given. We fix a step function f given as a finite linear combination of

characteristic functions of disjoint dyadic intervals. The class of such functions is

dense in all the Lp spaces. Once (5.3.13) is obtained for such functions, a density ar-

gument gives that T admits an extension on L1 that also satisfies (5.3.13). Therefore

it suffices to prove (5.3.13) for such a function f .

Apply the Calderon–Zygmund decomposition to f at height γα , where γ is a

positive constant to be chosen later. That is, write the function f as the sum

f = g+b = g+∑j

b j,

where conditions (1)–(6) of Theorem 5.3.1 are satisfied with the constant α replaced

by γα . Since f is a finite linear combination of characteristic functions of disjoint

dyadic cubes, there are only finitely many cubes Q j that appear in the Calderon–

Zygmund decomposition to f . Each b j is supported in a dyadic cube Q j with center

y j and the Q j’s are pairwise disjoint. We denote by ℓ(Q) the side length of a cube

Q. Let Q∗j be the unique cube with sides parallel to the axes having the same center

as Q j and having side length ℓ(Q∗j) = 2√

nℓ(Q j). Because of the form of f , each

b j is a bounded function supported in Q j, hence it is in Lr, thus each T (b j) is a

well-defined Lr function. We observe that for all j and all x /∈ Q∗j we have

T (b j)(x) = limk→∞

∫

k≥|x−y|≥δk

K(x− y)b j(y)dy =∫

Q j

K(x− y)b j(y)dy ,

where the last integral converges absolutely. This is a consequence of the Lebesgue

dominated convergence theorem, based on the facts that b j is bounded, that K is

integrable over any compact annulus that does not contain the origin (cf. (5.3.4)),

and that x−Q j is contained in such a compact annulus, since x /∈ Q∗j .Next we use the cancellation of b j in the following way:

∫

(∪iQ∗i )

c∑

j

∣∣T (b j)(x)∣∣ dx

=∫

(⋃

i Q∗i )c∑

j

∣∣∣∣∫

Q j

b j(y)(K(x− y)−K(x− y j)

)dy

∣∣∣∣dx

≤ ∑j

∫

(Q∗j )c

∫

Q j

|b j(y)||K(x− y)−K(x− y j)|dydx

= ∑j

∫

Q j

|b j(y)|∫

(Q∗j )c|K(x− y)−K(x− y j)|dxdy

= ∑j

∫

Q j

|b j(y)|∫

−y j+(Q∗j )c|K(x− (y− y j))−K(x)|dxdy

≤ ∑j

∫

Q j

|b j(y)|∫

|x|≥2|y−y j||K(x− (y− y j))−K(x)|dxdy

≤ A2∑j

∥∥b j

∥∥L1

≤ A22n+1∥∥ f

∥∥L1 < ∞ ,

where we used (5.3.12) since if x ∈ −y j + (Q∗j)c then |x| ≥ 1

2ℓ(Q∗j) =

√nℓ(Q j)

and since y− y j ∈ −y j +Q j we have |y− y j| ≤√

n

2ℓ(Q j), thus |x| ≥ 2|y− y j|. See

Figure 5.2.

Fig. 5.2 The cubes −y j +Q j

and −y j +Q∗j .

0.

.x

j*

j− y + Q

j j− y + Q

.j

y − y

Thus we proved that

∫

(⋃

i Q∗i )c∑

j

|T (b j)(x)|dx≤ 2n+1A2

∥∥ f∥∥

L1 ,

an inequality we use below. We have

∣∣x ∈ Rn : |T ( f )(x)|> α∣∣

≤∣∣∣

x ∈ Rn : |T (g)(x)|> α

2

∣∣∣+∣∣∣

x ∈ Rn : |T (b)(x)|> α

2

∣∣∣

≤ 2r

αr

∥∥T (g)∥∥r

Lr +∣∣∣⋃

i

Q∗i

∣∣∣+∣∣∣

x /∈⋃

i

Q∗i : |T(∑

j

b j

)(x)|> α

2

∣∣∣

=2r

αr

∥∥T (g)∥∥r

Lr +∣∣∣⋃

i

Q∗i

∣∣∣+∣∣∣

x /∈⋃

i

Q∗i : |∑j

T (b j)(x)|>α

2

∣∣∣

≤ 2r

αrBr

∥∥g∥∥r

Lr +∑i

|Q∗i |+2

α

∫

(⋃

i Q∗i )c∑

j

|T (b j)(x)|dx

≤ 2r

αr2

nrr′ Br(γα)

rr′∥∥ f

∥∥L1 +(2

√n)n

∥∥ f∥∥

L1

γα+

2

α2n+1A2

∥∥ f∥∥

L1

≤((2n+1Bγ)r

2nγ+

(2√

n)n

γ+2n+2A2

)∥∥ f∥∥

L1

α.

Choosing γ = 2−(n+1)B−1, we deduce the weak type (1,1) estimate (5.3.13) for T

with Cn = 2+2n+1(2√

n)n +2n+2.

It follows from Exercise 1.3.2 that T has an extension that maps Lp to Lp with

bound at most C′n(A2 +B)(p− 1)−1/p when 1 < p < r. The adjoint operator T ∗ of

T , defined by ⟨T ( f ) |g

⟩=

⟨f |T ∗(g)

⟩,

has a kernel that coincides with the function K∗(x) = K(−x) on Rn \0. We notice

that since K satisfies (5.3.12), then so does K∗ and with the same bound. Therefore,

T ∗, which maps Lr′ to Lr′ , has a kernel that satisfies (5.3.12). By the preceding

argument, T ∗ maps Lp′ to Lp′ with bound at most C′n(A2 +B)(p−1)−1/p whenever

1 < p′ < r′. By duality this yields that T maps Lp(Rn) to Lp(Rn) with bound at most

C′n(A2 +B)(p−1)1−1/p whenever r′ < p < ∞. Using interpolation we obtain that T

maps Lp to itself with norm at most C′n(A2 +B)max((p−1)−1/p,(p−1)1−1/p) for

p in the interval (r,r′), which is nonempty only if r < 2. Then (5.3.14) holds since

max((p−1)−1/p,(p−1)1−1/p)≤max((p−1)−1, p).

5.3.4 Discussion on Maximal Singular Integrals

In this subsection we introduce maximal singular integrals and we derive their

boundedness under certain smoothness conditions on the kernels, assuming bound-

edness of the associated linear operator.

Suppose that K is a kernel on Rn \0 that satisfies the size condition

|K(x)| ≤ A1|x|−n (5.3.15)

for x = 0. Then for any ε > 0 the function K(ε)(x) = K(x)χ|x|≥ε lies in Lp′(Rn)

(with norm bounded by cp,n A1ε−n/p) for all 1≤ p < ∞. Consequently, by Holder’s

inequality, the integral

( f ∗K(ε))(x) =∫

|y|≥εf (x− y)K(y)dy

converges absolutely for all x ∈ Rn and all f ∈ Lp(Rn), when 1≤ p < ∞.

Let f ∈⋃1≤p<∞Lp(Rn). We define the truncated singular integrals T (ε)( f ) asso-

ciated with the kernel K by setting

T (ε)( f ) = f ∗K(ε) ;

we also define the maximal truncated singular integral operator associated with K

by setting

T (∗)( f ) = supε>0

|( f ∗K(ε))|= supε>0

|T (ε)( f )| .

This operator is well defined, but possibly infinite, for certain points in Rn.

We now consider the situation in which the kernel K satisfies an integrability

condition over concentric annuli centered at the origin, a condition that is certainly

a weaker condition than (5.3.15). Precisely, suppose that K is a measurable function

on Rn \ 0, that is integrable on compact subsets of Rn \ 0, for which there is a

constant A1 < ∞ such that

supR>0

∫

R≤|x|≤2R|K(x)|dx≤ A1 < ∞ . (5.3.16)

Such kernels may not be integrable to the power p′ > 1 over the region |x| ≥ ε .

For this reason, it is not possible to define T (ε) as an absolutely convergent integral.

To overcome this difficulty, we consider double truncations. We define the doubly

truncated kernel K(ε ,N) by setting

K(ε ,N)(x) = K(x)χε≤|x|≤N(x) . (5.3.17)

A repeated application of (5.3.16) yields that

∫|K(ε ,N)(x)|dx≤ A1

([log2

N

ε

]+1

),

which implies that K(ε ,N) is integrable over concentric annuli centered at the origin.

Next, we define the doubly truncated singular integrals T (ε ,N) by setting

T (ε ,N)( f ) = f ∗K(ε ,N) ,

and we observe that these operators are well defined when f in Lp, for 1 ≤ p ≤ ∞.

Indeed, Theorem 1.2.10 yields that

∥∥T (ε ,N)( f )∥∥

Lp ≤∥∥ f

∥∥Lp

∫|K(ε ,N)(x)|dx < ∞

for functions f in Lp, 1≤ p≤ ∞. Consequently, for almost every x ∈ Rn we have

|T (ε ,N)( f )(x)|< ∞ .

For functions in⋃

1≤p≤∞Lp(Rn) we define the doubly truncated maximal singular

integral operator T (∗∗) associated with K by setting

T (∗∗)( f ) = sup0<ε<N<∞

|T (ε ,N)( f )| . (5.3.18)

For such functions and for almost all x ∈ Rn, T (∗∗)( f )(x) is well defined, but poten-

tially infinite.

One observation is that under condition (5.3.16), one can also define T (∗)(g)for general integrable functions g with compact support. In this case, say that the

ball B(0,R) contains the support of g. Let x ∈ B(0,M) and N = M + R. Then

|T (ε)(g)(x)| ≤ |g| ∗ |K(ε ,N)|(x), which is finite a.e. as the convolution of two L1

functions; consequently, the integral defining T (ε)(g)(x) converges absolutely for

all x ∈ B(0,R). Since R > 0 is arbitrary, T (ε)(g)(x) is defined and finite for almost

all x ∈ Rn.

Obviously T (∗) and T (∗∗) are related. If K satisfies condition (5.3.15), then

∣∣∣∣∫

ε≤|y|f (x− y)K(y)dy

∣∣∣∣≤ supN>0

∣∣∣∣∫

ε≤|y|≤Nf (x− y)K(y)dy

∣∣∣∣ ,

which implies that

T (∗)( f )≤ T (∗∗)( f )

for all f ∈⋃1≤p<∞Lp. Also, T (ε ,N)( f ) = T (ε)( f )−T (N)( f ); hence

T (∗∗)( f )≤ 2T (∗)( f ) .

Therefore, for kernels satisfying (5.3.15), T (∗∗) and T (∗) are comparable and the

boudnedness properties of T (∗∗) and T (∗) are equivalent

Theorem 5.3.4. (Cotlar’s inequality) Let 0 < A1,A2,A3 < ∞ and suppose that K is

defined on Rn \0 and satisfies the size condition,

|K(x)| ≤ A1|x|−n , x = 0, (5.3.19)

the smoothness condition

|K(x− y)−K(x)| ≤ A2|y|δ |x|−n−δ , (5.3.20)

whenever |x| ≥ 2|y|> 0, and the cancellation condition

sup0<r<R<∞

∣∣∣∣∫

r<|x|<RK(x)dx

∣∣∣∣≤ A3 . (5.3.21)

Let W in S ′(Rn) be related to K via (5.3.7) and let T be the operator given by

convolution with W. Then there is a constant Cn,δ such that the following inequality

is valid:

T (∗)( f )≤M(T ( f ))+Cn,δ (A1 +A2 +A3)M( f ) , (5.3.22)

for all f ∈ S (Rn), where M is the Hardy–Littlewood maximal operator. Conse-

quently, T (∗) is bounded on Lp(Rn) when 1 0 we use the notation gε(x) =ε−ng(ε−1x). For a distribution W we define Wε analogously, i.e. as the unique dis-

tribution with the property 〈Wε ,ψ〉 = ε−n〈W,ψε−1〉. We begin by observing that

Kε−1(x) = εnK(εx) satisfies (5.3.19), (5.3.20), and (5.3.21) uniformly in ε > 0.

Set, as before, K(ε)(x) = K(x)χ|x|≥ε . Fix f ∈S (Rn) for some 1 0:

∣∣((Kε−1)(1)−Wε−1 ∗ϕ)(x)

∣∣≤C(A1 +A2 +A3)(1+ |x|)−n−δ (5.3.24)

for all x ∈ Rn. Indeed, for |x| ≥ 1 we express the left-hand side in (5.3.24) as

∣∣∣∣∫

Rn

(Kε−1(x)−Kε−1(x− y)

)ϕ(y)dy

∣∣∣∣ .

Since ϕ is supported in |y| ≤ 1/2, we have |x| ≥ 2|y|, and condition (5.3.20) yields

that the expression on the left-hand side of (5.3.24) is bounded by

A2

|x|n+δ∫

Rn|y|δ |ϕ(y)|dy≤ c

A2

(1+ |x|)n+δ,

which proves (5.3.24) in the case |x| ≥ 1. When |x|< 1, the left-hand side of (5.3.24)

equals

(Wε−1 ∗ϕ)(x) = limδ j→0

∫

|x−y|≥δ j

Kε−1(x− y)ϕ(y)dy (5.3.25)

for some sequence δ j ↓ 0; see the discussion in Section 5.3.2. The expression in

(5.3.25) is equal to

I1 + I2 + I3 ,

where

In I1 we have 1/8 ≤ |x− y| ≤ 1+ 1/2 = 3/2; hence I1 is bounded by a multiple of

A1. Since |ϕ(x)−ϕ(y)| ≤ c|x− y|, the same is valid for I2. Finally, I3 is bounded

by a multiple of A3. Combining these facts yields the proof of (5.3.24) in the case

|x|< 1 as well.

I1 =

∫

|x−y|> 1

8

Kε−1(x− y)ϕ(y)dy ,

I2 =∫

|x−y|≤ 1

8

Kε−1(x− y)(

ϕ(y)−ϕ(x))

dy ,

I3 = ϕ(x) limδ j→0

∫

1

8≥|x−y|≥δ j

Kε−1(x− y)dy .

Use Corollary 2.1.12 to deduce that

supε>0

∣∣ f ∗((Kε−1)(1)−Kε−1 ∗ϕ

)ε

∣∣≤ c(A1 +A2 +A3)M( f ) .

Finally, take the supremum over ε > 0 in (5.3.23) and use (5.3.24) and Corollary

2.1.12 one more time to deduce the estimate

T (∗)( f )≤M( f ∗W )+C (A1 +A2 +A3)M( f ) ,

where C depends on n and δ ; this concludes the proof of (5.3.22) for all functions

f ∈S (Rn). Thus T (∗) is bounded on Lp, 1 < p < ∞, when restricted to Schwartz

functions.

Now given a general function g in Lp(Rn) we find a sequence h j in S (Rn) such

that ‖h j−g‖Lp → 0 as j→ ∞. Then we have the pointwise estimate

|T (ε)(g)| ≤ |T (ε)(g−h j)|+ |T (ε)(h j)| ≤ cp,n A1ε− n

p

0 ‖g−h j‖Lp + |T (∗)(h j)|

for all ε ≥ ε0. Taking the supremum over ε ≥ ε0 and then Lp norm over the ball

B(0,R), we obtain

∥∥ supε≥ε0

|T (ε)(g)|∥∥

Lp(B(0,R))≤ c′p,n A1ε

− np

0 Rnp ‖g−h j‖Lp +C′ (A1 +A2 +A3)‖h j‖Lp .

Now we let j → ∞ first, and then R→ ∞ and ε0 → 0 to deduce the boundedness of

T (∗) on Lp(Rn) via the Lebesgue monotone convergence theorem.

5.3.5 Boundedness for Maximal Singular Integrals Implies

Weak Type (1,1) Boundedness

We now state and prove a result analogous to that in Theorem 5.3.3 for maximal

singular integrals.

Theorem 5.3.5. Let K(x) be function on Rn \ 0 satisfying (5.3.4) with constant

A1 <∞ and Hormander’s condition (5.3.12) with constant A2 <∞. Suppose that the

operator T (∗∗) as defined in (5.3.18) maps L2(Rn) to itself with norm B. Then T (∗∗)

maps L1(Rn) to L1,∞(Rn) with norm

∥∥T (∗∗)∥∥L1→L1,∞ ≤Cn(A1 +A2 +B),

where Cn is some dimensional constant.

Proof. The proof of this theorem is only a little more involved than the proof of

Theorem 5.3.3. We fix an L1(Rn) function f . We apply the Calderon–Zygmund

decomposition of f at height γα for some γ ,α > 0. We then write f = g+b, where

b = ∑ j b j and each b j is supported in some cube Q j. We define Q∗j as the cube

with the same center as Q j and with sides parallel to the sides of Q j having length

ℓ(Q∗j) = 5√

nℓ(Q j). This is only a minor change compared with the definition of Q j

in Theorem 5.3.3. The main change in the proof is in the treatment of the term

∣∣∣

x ∈(⋃

j

Q∗j)c

: |T (∗∗)(b)(x)|> α

2

∣∣∣ . (5.3.26)

We show that for all γ ≤ (2n+5A1)−1 we have

∣∣∣

x ∈(⋃

j

Q∗j)c

: |T (∗∗)(b)(x)|> α

2

∣∣∣≤ 2n+8A2

∥∥ f∥∥

L1

α. (5.3.27)

Let us conclude the proof of the theorem assuming for the moment the validity of

(5.3.27). As in the proof of Theorem 5.3.3, we can show that

∣∣∣

x ∈ Rn : |T (∗∗)(g)(x)|> α

2

∣∣∣+∣∣∣⋃

j

Q∗j

∣∣∣≤(

2n+2B2γ+(5√

n)n

γ

)∥∥ f∥∥

L1

α.

Combining this estimate with (5.3.27) and choosing

γ = (2n+5(A1 +A2 +B))−1 ,

we obtain the required estimate

∣∣x ∈ Rn : |T (∗∗)( f )(x)|> α∣∣≤Cn(A1 +A2 +B)

∥∥ f∥∥

L1

α

with Cn = 2−3 +(5√

n)n2n+5 +2n+8.

It remains to prove (5.3.27). This estimate will be a consequence of the fact that

for x ∈(⋃

j Q∗j)c

we have the key inequality

T (∗∗)(b)(x)≤ 4E1(x)+2n+2αγE2(x)+2n+3αγA1 , (5.3.28)

where

E1(x) = ∑j

∫

Q j

|K(x− y)−K(x− y j)| |b j(y)|dy ,

E2(x) = ∑j

∫

Q j

|K(x− y)−K(x− y j)|dy ,

and y j is the center of Q j.

If we had (5.3.28), then we could easily derive (5.3.27). Indeed, fix a γ satisfying

γ ≤ (2n+5A1)−1. Then we have 2n+3αγA1 <

α3

, and using (5.3.28), we obtain


∣∣∣

x ∈(⋃

j

Q∗j)c

: |T (∗∗)(b)(x)|> α

2

∣∣∣

≤∣∣∣

x ∈(⋃

j

Q∗j)c

: 4E1(x)>α

12

∣∣∣

+∣∣∣

x ∈(⋃

j

Q∗j)c

: 2n+2αγE2(x)>α

12

∣∣∣

≤ 48

α

∫

(⋃

j Q∗j )cE1(x)dx+2n+6γ

∫

(⋃

j Q∗j )cE2(x)dx ,

(5.3.29)

since α2= α

3+ α

12+ α

12. We have

∫

(⋃

j Q∗j )cE1(x)dx

≤ ∑j

∫

Q j

|b j(y)|∫

(Q∗j )c|K(x− y)−K(x− y j)|dxdy

≤ ∑j

∫

Q j

|b j(y)|∫

|x−y j |≥2|y−y j||K(x− y)−K(x− y j)|dxdy

≤ A2∑j

∫

Q j

|b j(y)|dy = A2∑j

∥∥b j

∥∥L1 ≤ A22n+1

∥∥ f∥∥

L1 ,

(5.3.30)

where we used the fact that if x ∈ (Q∗j)c, then |x− y j| ≥ 1

2ℓ(Q∗j) =

52

√nℓ(Q j). But

since |y−y j| ≤√

n

2ℓ(Q j), this implies that |x−y j| ≥ 2|y−y j|. Here we used the fact

that the diameter of a cube is equal to√

n times its side length. Likewise, we obtain

that ∫

(⋃

j Q∗j )cE2(x)dx≤ A2∑

j

|Q j| ≤ A2

∥∥ f∥∥

L1

αγ. (5.3.31)

Combining (5.3.30) and (5.3.31) with (5.3.29) yields (5.3.27).

Fig. 5.3 The cubes Q j

and Q∗j .

x

y

y0

Qj

Qj*

j.

.

.

Therefore, the main task of the proof is to prove (5.3.28). Since b = ∑ j b j, to

estimate T (∗∗)(b), it suffices to estimate each |T (ε ,N)(b j)| uniformly in ε and N. To

achieve this we use the estimate

|T (ε ,N)(b j)| ≤ |T (ε)(b j)|+ |T (N)(b j)| , (5.3.32)

noting that the truncated singular integrals T (ε)(b j) are well defined. Indeed, say x

lies in a compact set K0. Pick M such that K0−Q j is contained in a ball B(0,M).Then

|T (ε)(b j)(x)| ≤ |b j| ∗ |K(ε ,M)|(x) ,which is finite a.e. as the convolution of two L1 functions; thus the integral defining

T (ε)(b j)(x) converges absolutely and the expression T (ε)(b j)(x) is well defined for

almost all x.

We work with T (ε) and we note that T (N) can be treated similarly. Fix x /∈ ⋃j Q∗j

and ε > 0 and define

J1(x,ε) = j : ∀y ∈ Q j we have |x− y|< ε,J2(x,ε) = j : ∀y ∈ Q j we have |x− y|> ε,J3(x,ε) = j : ∃y ∈ Q j we have |x− y|= ε.

Note that

T (ε)(b j)(x) = 0

whenever x /∈⋃j Q∗j and j ∈ J1(x,ε). Also note that

K(ε)(x− y) = K(x− y)

whenever x /∈⋃j Q∗j , j ∈ J2(x,ε) and y ∈ Q j. Therefore,

supε>0

|T (ε)(b)(x)| ≤ supε>0

∣∣ ∑j∈J2(x,ε)

T (b j)(x)∣∣+ sup

ε>0

∣∣ ∑j∈J3(x,ε)

T (b jχ|x−·|≥ε)(x)∣∣ ,

but since

supε>0

∣∣ ∑j∈J2(x,ε)

T (b j)(x)∣∣≤∑

j

|T (b j)(x)| ≤ E1(x) , (5.3.33)

it suffices to estimate the term

supε>0

∣∣∣ ∑j∈J3(x,ε)

T (b jχ|x−·|≥ε)(x)∣∣∣ .

We now make some geometric observations; see Figure 5.3. Fix ε > 0 and a cube

Q j with j ∈ J3(x,ε); recall that x lies in (⋃

j Q∗j)c. Then we have

ε ≥ 1

2

(ℓ(Q∗j)− ℓ(Q j)

)=

1

2(5√

n−1)ℓ(Q j)≥ 2√

nℓ(Q j) . (5.3.34)


Since j ∈ J3(x,ε), there exists a y0 ∈ Q j with

|x− y0|= ε .

Using (5.3.34), we obtain that for any y ∈ Q j we have

ε

2≤ ε−

√nℓ(Q j)≤ |x− y0|− |y− y0| ≤ |x− y| ,

|x− y| ≤ |x− y0|+ |y− y0| ≤ ε+√

nℓ(Q j)≤3ε

2.

Therefore, we have proved that

⋃

j∈J3(x,ε)

Q j B(x, 3ε2)\B(x, ε

2) .

Letting

c j(ε) =1

|Q j|

∫

Q j

b j(y)χ|x−y|≥ε(y)dy ,

we note that in view of property (5) of the Calderon–Zygmund decomposition (The-

orem 5.3.1), the estimate

|c j(ε)| ≤ 2n+1αγ

holds. Then

supε>0

∣∣∣∣ ∑j∈J3(x,ε)

∫

Q j

K(x− y)b j(y)χ|x−y|≥ε(y)dy

∣∣∣∣

≤ supε>0

∣∣∣∣ ∑j∈J3(x,ε)

∫

Q j

K(x− y)(b j(y)χ|x−y|≥ε(y)− c j(ε)

)dy

∣∣∣∣

+ supε>0

∣∣∣∣ ∑j∈J3(x,ε)

c j(ε)∫

Q j

K(x− y)dy

∣∣∣∣

≤ supε>0

∣∣∣∣ ∑j∈J3(x,ε)

∫

Q j

(K(x− y)−K(x− y j)

)(b j(y)χ|x−y|≥ε(y)− c j(ε)

)dy

∣∣∣∣

+2n+1αγ supε>0

∫

B(x, 3ε2 )\B(x, ε2 )

|K(x− y)|dy

≤ ∑j

∫

Q j

∣∣K(x− y)−K(x− y j)∣∣(|b j(y)|+2n+1αγ

)dy

+2n+1αγ supε>0

∫

ε2≤|x−y|≤ 3ε

2

|K(x− y)|dy

≤ E1(x)+2n+1αγE2(x)+2n+1αγ(2A1) .

The last estimate, together with (5.3.33), with (5.3.32), and with the analogous esti-

mate for supN>0 |T (N)(b j)(x)| (which is similarly obtained), yields (5.3.28).

The value of the previous theorem lies in the following: Since we know that for

some sequences ε j ↓ 0, N j ↑∞ the pointwise limit T (ε j ,N j)( f ) exists a.e. for all f in a

dense subclass of L1, then Theorem 5.3.5 allows us to deduce that T (ε j ,N j)( f ) exists

a.e. for all f in L1(Rn).If the singular integrals have kernels of the form Ω(x/|x|)|x|−n with Ω in L∞,

such as the Hilbert transform and the Riesz transforms, then the upper truncations

are not needed for K in (5.3.17). In this case

T(ε)Ω ( f )(x) =

∫

|y|≥εf (x− y)

Ω(y/|y|)|y|n dy

is well defined for f ∈⋃1≤p<∞Lp(Rn) by Holder’s inequality and is equal to

limN→∞

∫


Ω(y/|y|)|y|n dy .

Corollary 5.3.6. The maximal Hilbert transform H(∗) and the maximal Riesz trans-

forms R(∗)j are weak type (1,1). Secondly, limε→0 H(ε)( f ) and limε→0 R

(ε)j (g) exist

a.e. for all f ∈ L1(R) and g ∈ L1(Rn), as ε → 0.

Proof. Since the kernels 1/x on R and x j/|x|n on Rn satisfy (5.3.10), the first state-

ment in the corollary is an immediate consequence of Theorem 5.3.5. The second

statement follows from Theorem 2.1.14 and Corollary 5.2.8, since these limits exist

for Schwartz functions.

Corollary 5.3.7. Under the hypotheses of Theorem 5.3.5, T (∗∗) maps Lp(Rn) to itself

for 1 0

1

R

∫

|x|≤R|K(x)| |x|dx≤ 2A1 ;

thus the expressions in (5.3.6) and (5.3.4) are equivalent.

5.3.2. Suppose that K is a locally integrable function on Rn \ 0 that satisfies

(5.3.4). Suppose that δ j ↓ 0. Prove that the principal value operation


∫

δ j≤|x|≤1K(x)ϕ(x)dx

defines a distribution in S ′(Rn) if and only if the following limit exists:

limj→∞

∫

δ j≤|x|≤1K(x)dx .

5.3.3. Suppose that a function K on Rn \ 0 satisfies condition (5.3.4) with con-

stant A1 and condition (5.3.12) with constant A2.

(a) Show that the functions K(x)χ|x|≥ε also satisfy condition (5.3.12) uniformly in

ε > 0 with constant A1 +A2.

(b) Obtain the same conclusion for the upper truncations K(x)χ|x|≤N.

(c) Deduce a similar conclusion for the double truncations K(ε ,N)(x)=K(x)χε≤|x|≤N .

5.3.4. Modify the proof of Theorem 5.3.5 to prove that if T (∗∗) maps Lr to Lr,∞ for

some 1 < r < ∞, and K satisfies condition (5.3.12), then T (∗∗) maps L1 to L1,∞.

5.3.5. Assume that T is a linear operator acting on measurable functions on Rn

such that whenever a function f is supported in a cube Q, then T ( f ) is supported in

a fixed multiple of Q.

(a) Suppose that T maps Lp to itself for some 1 < p < ∞ with norm B. Prove that T

extends to a bounded operator from L1 to L1,∞ with norm a constant multiple of B.

(b) Suppose that T maps Lp to Lq for some 1 < q < p < ∞ with norm B. Prove that

T extends to a bounded operator from L1 to Ls,∞ with norm a multiple of B, where

1

p′+

1

q=

1

s.

5.3.6. (a) Let 1 < q <∞. Show that the good function g in Theorem 5.3.1 lies in the

Lorentz space Lq,1 and that ‖g‖Lq,1 ≤Cn,qα1/q′‖ f‖1/q

L1 for some constant Cn,q.

(b) Let 1< r <∞. Obtain a generalization of Theorem 5.3.3 in which the assumption

that T maps Lr to Lr is replaced by that T maps Lr,1 to Lr,∞ with norm B.

(c) Let 1 < r < ∞. Obtain a further generalization of Theorem 5.3.3 in which the

assumption that T maps Lr to Lr is replaced by that it is of restricted weak type

(r,r), i.e., it satisfies

|x : |T (χE)(x)|> α| ≤ Br |E|αr

for all subsets E of Rn with finite measure.

5.3.7. Let K satisfy (5.3.12) for some A2 > 0, let W ∈S ′(Rn) be an extension of K

on Rn as in (5.3.7), and let T be the operator given by convolution with W . Obtain

the case r = ∞ in Theorem 5.3.3. Precisely, prove that if T maps L∞(Rn) to itself

with constant B, then T has an extension on L1 +L∞ that satisfies

∥∥T∥∥

L1→L1,∞ ≤C′n (A2 +B),

and for 1 < p < ∞ it satisfies

∥∥T∥∥

Lp→Lp ≤Cn

1

(p−1)1/p(A2 +B),

where Cn,C′n are constants that depend only on the dimension.[

Hint: Apply the Calderon–Zygmund decomposition f = g+b at height αγ , where

γ = (2n+1B)−1. Since |g| ≤ 2nαγ , observe that

|x : |T ( f )(x)|> α| ≤ |x : |T (b)(x)|> α/2| .

For the interpolation use the result of Exercise 1.3.2.]

5.3.8. (Calderon–Zygmund decomposition on Lq) Fix a function f ∈ Lq(Rn) for

some 1≤ q < ∞ and let α > 0. Then there exist functions g and b on Rn such that

(1) f = g+b.

(2) ‖g‖Lq ≤ ‖ f‖Lq and ‖g‖L∞ ≤ 2nqα .

(3) b = ∑ j b j, where each b j is supported in a cube Q j. Furthermore, the cubes Qk

and Q j have disjoint interiors when j = k.

(4) ‖b j‖qLq ≤ 2n+qαq|Q j|.

(5)∫

Q jb j(x)dx = 0.

(6) ∑ j |Q j| ≤ α−q‖ f‖qLq .

(7) ‖b‖Lq ≤ 2n+q

q ‖ f‖Lq and ‖b‖L1 ≤ 2α1−q‖ f‖qLq .

[Hint: Imitate the basic idea of the proof of Theorem 5.3.1, but select a cube Q if(

1|Q|

∫Q | f (x)|q dx

)1/q> α . Define g and b as in the proof of Theorem 5.3.1.

]

5.3.9. Let f ∈ L1(Rn). Then for any α > 0, prove that there exist disjoint cubes Q j

in Rn such that the set Eα = x ∈ Rn : Mc( f )(x) > α is contained in⋃

j 3Q j andα4n < 1

|Q j |∫

Q j| f (t)|dt ≤ α

2n .[Hint: For given α > 0, select all maximal dyadic cubes Q j(α) such that the average

of f over them is bigger than α . Given x ∈ Eα , pick a cube R that contains x such

that the average of | f | over R is bigger than α and find a dyadic cube Q such that

2−n|Q| < |R| ≤ |Q| and that∫

R∩Q | f |dx > 2−nα|R|. Conclude that Q is contained

in some Qk(4−nα) and thus R is contained in 3Qk(4

−nα). The collection of all

Q j = Q j(4−nα) is the required one.

]

5.3.10. Let K(x) be a function on Rn \0 that satisfies |K(x)| ≤ A|x|−n. Let η(x)be a smooth function equal to 1 when |x| ≥ 2 and vanishing when |x| ≤ 1. For f ∈ Lp,

1≤ p < ∞, define truncated singular integral operators

T (ε)( f )(x) =∫

|y|≥εK(y) f (x− y)dy,

T(ε)η ( f )(x) =

∫

Rnη(y/ε)K(y) f (x− y)dy .

Show that the truncated maximal singular integral T (∗)( f ) = supε>0 |T (ε)( f )| is Lp

bounded for 1 0 |T

(ε)η ( f )| is Lp bounded. Formulate an analogous statement

for p = 1.

5.3.11. (M. Mastyło) Let 1 ≤ p < ∞. Suppose that Tε are linear operators defined

on Lp(Rn) such that for all f ∈ Lp(Rn) we have |Tε( f )| ≤ Aε−a‖ f‖Lp for some

0 < a,A < ∞. Also suppose that there is a constant C < ∞ such that the maximal

operator T∗( f ) = supε>0 |Tε( f )| satisfies∥∥T∗(h)‖Lp ≤ C‖h‖Lp for all h ∈ S (Rn).

Prove that the same inequality is valid for all f ∈ Lp(Rn).[Hint: For a fixed δ > 0 define Sδ ( f ) = supε>δ |Tε( f )|, which is a subadditive func-

tional on Lp(Rn). For a fixed f0 ∈ Lp(Rn) define a linear space X0 = λ f0 : λ ∈Cand a linear functional T0 on X0 by setting T0(λ f0) = λSδ ( f0). By the Hahn–

Banach theorem there is an extension T0 of T0 that satisfies |T0( f )| ≤ Sδ ( f ) for

all f ∈ Lp(Rn). Since Sδ is Lp is bounded on Schwartz functions with norm at most

C, then so is T0. But T0 is linear and by density it is bounded on Lp(Rn) with norm

at most C; consequently, ‖Sδ ( f0)‖Lp = ‖T0( f0)‖Lp = ‖T0( f0)‖Lp ≤ C‖ f0‖Lp . The

required conclusion for T∗ follows by Fatou’s lemma.]

5.4 Sufficient Conditions for Lp Boundedness

We have used the Calderon–Zygmund decomposition to prove weak type (1,1)boundedness for singular integral and maximal singular integral operators, assum-

ing that these operators are already L2 bounded. It is therefore natural to ask for

sufficient conditions that imply L2 boundedness for such operators. Precisely, what

are sufficient conditions on functions K on Rn \ 0 so that the corresponding sin-

gular and maximal singular integral operators associated with K are L2 bounded?

We saw in Section 5.2 that if K has the special form K(x) =Ω(x/|x|)/|x|n for some

Ω ∈ L1(Sn−1) with mean value zero, then condition (5.2.16) is necessary and suf-

ficient for the L2 boundedness of T , while the L2 boundedness of T (∗) requires the

stronger smoothness condition (5.2.24).

For the general K considered in this section (for which the corresponding op-

erator does not necessarily commute with dilations), we only give some sufficient

conditions for L2 boundedness of T and T (∗∗).Throughout this section K denotes a locally integrable function on Rn \0 that

satisfies the “size” condition

supR>0

∫

R≤|x|≤2R|K(x)|dx = A1 < ∞ , (5.4.1)

5.4 Sufficient Conditions for Lp Boundedness 375

the “smoothness” condition

supy =0

∫

|x|≥2|y||K(x− y)−K(x)|dx = A2 < ∞ , (5.4.2)

and the “cancellation” condition

sup0<R1<R2<∞

∣∣∣∣∫

R1<|x|<R2

K(x)dx

∣∣∣∣= A3 < ∞ , (5.4.3)

for some A1,A2,A3 > 0. As mentioned earlier, condition (5.4.2) is often referred to

as Hormander’s condition. In this section we show that these three conditions give

rise to convolution operators that are bounded on Lp.

5.4.1 Sufficient Conditions for Lp Boundedness of Singular

Integrals

We first note that under conditions (5.4.1), (5.4.2), and (5.4.3), there exists a tem-

pered distribution W of the form (5.3.7) that coincides with K on Rn \0. Indeed,

condition (5.4.3) implies that there exists a sequence δ j ↓ 0 such that

limj→∞

∫

δ j<|x|≤1K(x)dx = L

exists. Using (5.3.8), we conclude that there exists such a tempered distribution W .

Note that we must have |L| ≤ A3.

We observe that the difference of two distributions W and W ′ that coincide with

K on Rn \0 must be supported at the origin.

Theorem 5.4.1. Assume that K satisfies (5.4.1), (5.4.2), and (5.4.3), and let W be a

tempered distribution of the form (5.3.7) that coincides with K on Rn \0. Then we

have

sup0<ε<N<∞

supξ∈Rn

|(Kχε<| · |<N)∧(ξ )| ≤ 15(A1 +A2 +A3) (5.4.4)

and consequently

supξ∈Rn

|W (ξ )| ≤ 15(A1 +A2 +A3) . (5.4.5)

Thus the operator given by convolution with W maps L2(Rn) to itself with norm at

most 15(A1 +A2 +A3). Consequently, it also maps L1(Rn) to L1,∞(Rn) with bound

at most a dimensional constant multiple of A1 +A2 +A3 and Lp(Rn) to itself with

bound at most Cn max(p,(p−1)−1)(A1 +A2 +A3), when 1 < p < ∞, where Cn is a

dimensional constant.

Proof. Let us set K(ε ,N)(x) = K(x)χε<|x|<N . Estimate (5.4.4) implies that for all f

in S (Rn) we have

∥

∥ f ∗K(δ j , j)∥

∥

L2 ≤ 15(A1 +A2 +A3)∥

∥ f∥

∥

L2

uniformly in j. Using this, (5.3.9), and Fatou’s lemma, we obtain that

∥

∥ f ∗W∥

∥

L2 ≤ 15(A1 +A2 +A3)∥

∥ f∥

∥

L2 ,

for all f ∈ S (Rn), and this is equivalent to (5.4.5).

Case 1: Suppose that ε < |ξ |−1 < N. Then we write

K(ε ,N)(ξ ) = I1(ξ )+ I2(ξ ),

where

I1(ξ ) =∫

ε<|x|<|ξ |−1K(x)e−2πix·ξ dx ,

I2(ξ ) =∫

|ξ |−1<|x|<NK(x)e−2πix·ξ dx .

We now have

I1(ξ ) =∫

ε<|x|<|ξ |−1K(x)dx+

∫

ε<|x|<|ξ |−1K(x)(e−2πix·ξ −1)dx. (5.4.6)

It follows that

|I1(ξ )| ≤ A3 +2π|ξ |∫

|x|<|ξ |−1|x| |K(x)|dx ≤ A3 +2π(2A1)

uniformly in ε . Let us now examine I2(ξ ). Let z = ξ2|ξ |2

so that e2πiz·ξ = −1 and

2|z|= |ξ |−1. By changing variables x = x′− z, rewrite I2 as

I2(ξ ) =−∫

|ξ |−1<|x′−z|<NK(x′− z)e−2πix′·ξ dx′ ;

hence averaging gives

I2(ξ ) =1

2

∫

|ξ |−1<|x|<NK(x)e−2πix·ξ dx−

1

2

∫

|ξ |−1<|x−z|<NK(x− z)e−2πix·ξ dx .

Now use that∫

AF dx−

∫

BGdx =

∫

B(F −G)dx+

∫

A\BF dx−

∫

B\AF dx (5.4.7)

to write I2(ξ ) = J1(ξ )+ J2(ξ )+ J3(ξ )+ J4(ξ )+ J5(ξ ), where

J1(ξ ) = +1

2

∫

|ξ |−1<|x−z|<N

(K(x)−K(x− z)

)e−2πix·ξ dx , (5.4.8)

J2(ξ ) = +1

2

∫

|ξ |−1<|x|<N

|x−z|≤|ξ |−1

K(x)e−2πix·ξ dx , (5.4.9)

J3(ξ ) = +1

2

∫

|ξ |−1<|x|<N

|x−z|≥N

K(x)e−2πix·ξ dx , (5.4.10)

J4(ξ ) = − 1

2

∫

|ξ |−1<|x−z|<N

|x|≤|ξ |−1

K(x)e−2πix·ξ dx , (5.4.11)

J5(ξ ) = − 1

2

∫

|ξ |−1<|x−z|<N

|x|≥N

K(x)e−2πix·ξ dx . (5.4.12)

Since 2|z|= |ξ |−1, J1(ξ ) is bounded in absolute value by 12A2, in view of (5.4.2).

Next observe that |ξ |−1 ≤ |x| ≤ 32|ξ |−1 in (5.4.9), while 1

2|ξ |−1 ≤ |x| ≤ |ξ |−1 in

(5.4.11); hence both J2 and J4 are bounded by 12A1. Finally, we have 1

2N < |x| < N

in (5.4.10) (since |x|> N− 12|ξ |−1), and similarly we have N ≤ |x|< 3

2N in (5.4.12).

Thus both J3 and J5 are bounded above by 12A1.

Case 2: If ε < N ≤ |ξ |−1, then we write

∫

ε<|x|<NK(x)e−2πix·ξ dx =

∫

ε<|x|<NK(x)dx+

∫

ε<|x|<NK(x)(e−2πix·ξ −1)dx

which is bounded in absolute value by

A3 +2π|ξ |∫

|x|≤|ξ |−1|K(x)| |x|dx≤ A3 +4πA1.

Notice that if ξ = 0, only the first term appears which is bounded by A3.

Case 2: If |ξ |−1 ≤ ε < N, we write

∫

ε<|x|<NK(x)e−2πix·ξdx =

∫

|ξ |−1<|x|<NK(x)e−2πix·ξdx−

∫

|ξ |−1<|x|<εK(x)e−2πix·ξdx,

and both of the terms on the right are similar to I2(ξ ) which was shown to be

bounded by 2A1 +12A2.

In all cases (5.4.4) holds.

5.4.2 An Example

We now give an example of a distribution that satisfies conditions (5.4.1), (5.4.2),

and (5.4.3).

Example 5.4.2. Let τ be a nonzero real number and let K(x) = 1|x|n+iτ defined for

x ∈ Rn \0. For a sequence δk ↓ 0 and ϕ a Schwartz function on Rn, define

⟨W,ϕ

⟩= lim

k→∞

∫

δk≤|x|ϕ(x)

dx

|x|n+iτ, (5.4.13)

whenever the limit exists. We claim that for some choices of sequences δk, W is a

well defined tempered distribution on Rn. Take, for example, δk = e−2πk/τ . For this

sequence δk, observe that

∫

δk≤|x|≤1

1

|x|n+iτdx = ωn−1

1− (e−2πk/τ)−iτ

−iτ= 0 ,

and thus

⟨W,ϕ

⟩=

∫

|x|≤1(ϕ(x)−ϕ(0)) dx

|x|n+iτ+

∫

|x|≥1ϕ(x)

dx

|x|n+iτ, (5.4.14)

which implies that W ∈S ′(Rn), since

∣∣⟨W,ϕ⟩∣∣≤C

[∥∥∇ϕ∥∥

L∞+

∥∥ |x|ϕ(x)∥∥

L∞

].

If ϕ is supported in Rn \0, then

⟨W,ϕ

⟩=

∫

RnK(x)ϕ(x)dx .

Therefore W coincides with the function K away from the origin. Moreover, (5.4.1)

and (5.4.2) are clearly satisfied for K, while (5.4.3) is also satisfied, since

∣∣∣∣∫

R1<|x|<R2

1

|x|n+iτdx

∣∣∣∣= ωn−1

∣∣∣∣∣R−iτ

1 −R−iτ2

−iτ

∣∣∣∣∣≤2ωn−1

|τ | .

Remark 5.4.3. It is important to emphasize that the limit in (5.4.13) may not exist

for all sequences δk → 0. For example, the limit in (5.4.13) does not exist if δk =e−πk/τ . Moreover, for a different choice of a sequence δk for which the limit in

(5.4.13) exists (for example, δk = e−π(2k+1)/τ ), we obtain a different distribution W1

that coincides with the function K(x) = |x|−n−iτ on Rn \0.We discuss a point of caution. We can directly check that the distributions W

defined by (5.4.13) are not homogeneous distributions of degree −n− iτ . In fact,

the only homogeneous distribution of degree −n− iτ that coincides with the func-

tion |x|−n−iτ away from zero is a multiple of the distribution u−n−iτ , where uz is

defined in (2.4.7). Let us investigate the relationship between u−n−iτ and W defined

in (5.4.14). Recall that (2.4.8) gives

⟨u−n−iτ ,ϕ

⟩=

∫

|x|≥1ϕ(x)

π−i τ2

Γ (−i τ2)|x|−n−iτ dx

+

∫

|x|≤1(ϕ(x)−ϕ(0)) π−i τ2

Γ (−i τ2)|x|−n−iτ dx+

ωn−1π−i τ2

−iτΓ (−i τ2)ϕ(0) .

Using (5.4.14), we conclude that u−n−iτ−c1W = c2δ0 for suitable nonzero constants

c1 and c2. Since the Dirac mass at the origin is not a homogeneous distribution of

degree −n− iτ , it follows that neither is W .

Since

u−n−iτ = uiτ = c3|ξ |iτ ,the identity u−n−iτ − c1W = c2δ0 can be used to obtain a formula for the Fourier

transform of W and thus produce a different proof that convolution with W is a

bounded operator on L2(Rn).

5.4.3 Necessity of the Cancellation Condition

Although conditions (5.4.1), (5.4.2), and (5.4.3) are sufficient for L2 boundedness,

they might not necessary. However, we show that (5.4.3) is a necessary condition,

given (5.4.1).

Proposition 5.4.4. Suppose that K is a function on Rn \ 0 that satisfies (5.4.1).

Let W be a tempered distribution on Rn related to K as in (5.3.7). If the operator

T given by convolution with W maps L2(Rn) to itself (equivalently if W is an L∞

function), then the function K must satisfy (5.4.3).

Proof. Pick a radial C ∞ function ϕ supported in the ball |x| ≤ 2 with 0 ≤ ϕ ≤ 1,

and ϕ(x) = 1 when |x| ≤ 1. For R > 0 let ϕR(x) = ϕ(x/R). Fourier inversion for

distributions gives the second equality,

(W ∗ϕR)(0) =⟨W,ϕR

⟩=

⟨W , ϕR

⟩=

∫

RnW (ξ )Rnϕ(Rξ )dξ ,

and the preceding identity implies that

|(W ∗ϕR)(0)| ≤∥∥W

∥∥L∞

∥∥ϕ∥∥

L1 =∥∥T

∥∥L2→L2

∥∥ϕ∥∥

L1

uniformly in R > 0. Fix 0 < R1 < R2 < ∞. If R2 ≤ 2R1, we have

∣∣∣∣∫

R1<|x|<R2

K(x)dx

∣∣∣∣≤∫

R1<|x|<2R1

|K(x)|dx≤ A1 ,

which implies the required conclusion. We may therefore assume that 2R1 < R2.

Since the part of the integral in (5.4.3) over the set R1 < |x| < 2R1 is controlled by

A1, it suffices to control the integral of K(x) over the set 2R1 < |x| < R2. Since the

function ϕR2 −ϕR1 is supported away from the origin, the action of the distribution

W on it can be written as integration against the function K. We have

∫

RnK(x)(ϕR2(x)−ϕR1(x))dx

=∫

2R1<|x|<R2

K(x)dx+∫

R1<|x|<2R1

K(x)(1−ϕR1(x))dx+∫

R2<|x|<2R2

K(x)ϕR2(x)dx.

The sum of the last two integrals is bounded by 3A1 (since 0 ≤ ϕ ≤ 1), while the

first integral is equal to

(W ∗ϕR2)(0)− (W ∗ϕR1)(0)

and is therefore bounded by 2‖T‖L2→L2‖ϕ ‖L1 . We conclude that the function K

must satisfy (5.4.3) with constant

A3 ≤ 3A1 +2‖ϕ ‖L1‖T‖L2→L2 ≤ c(A1 +‖T‖L2→L2

).

This establishes the assertion.

5.4.4 Sufficient Conditions for Lp Boundedness of Maximal

Singular Integrals

We now discuss the analogous result to Theorem 5.4.1 for the maximal singular

integral operator T (∗∗).

Theorem 5.4.5. Suppose that K satisfies (5.4.1), (5.4.2), and (5.4.3) and let T (∗∗) be

as in (5.3.18). Then T (∗∗) is bounded on Lp(Rn), 1 < p < ∞, with norm

∥∥T (∗∗)∥∥Lp→Lp ≤Cn max(p,(p−1)−1)(A1 +A2 +A3),

where Cn is a dimensional constant.

Proof. We first define an operator T associated with K that satisfies (5.4.1), (5.4.2),

and (5.4.3). Because of condition (5.4.3), there exists a sequence δ j ↓ 0 such that

limj→∞

∫

δ j<|x|≤1K(x)dx

exists. Therefore, for ϕ ∈S (Rn) we can define a tempered distribution

⟨W,ϕ

⟩= lim

j→∞

∫

δ j≤|x|≤ jK(x)ϕ(x)dx

and an operator T given by T ( f ) = f ∗W for f ∈S (Rn). In view of Theorems 5.4.1

and 5.3.3, T admits an Lp bounded extension (1 < p < ∞) with

∥∥T∥∥

Lp→Lp ≤ cn max(p,(p−1)−1)(A1 +A2 +A3) (5.4.15)

and is weak type (1,1). This extension is still denoted by T .

Fix 1 < p < ∞ and f ∈ Lp(Rn)∩L∞(Rn) with compact support. We have

T (ε ,N)( f )(x)

=∫

ε≤|x−y|<NK(x− y) f (y)dy = T (ε)( f )(x)−T (N)( f )(x)

=∫

ε≤|x−y|K(x− y) f (y)dy−

∫

N≤|x−y|K(x− y) f (y)dy

=∫

ε≤|x−y|(K(x− y)−K(z1− y)) f (y)dy+

∫

ε≤|x−y|K(z1− y) f (y)dy

−∫

N≤|x−y|(K(x− y)−K(z2− y)) f (y)dy−

∫

N≤|x−y|K(z2− y) f (y)dy

=∫

ε≤|x−y|(K(x− y)−K(z1− y)) f (y)dy+T ( f )(z1)−T ( f χ|x−·|<ε)(z1)

−∫

N≤|x−y|(K(x− y)−K(z2− y)) f (y)dy−T ( f )(z2)+T ( f χ|x−·|<N)(z2) ,

where z1 and z2 are arbitrary points in Rn that satisfy |z1− x| ≤ ε2

and |z2− x| ≤ N2

.

We used that f has compact support in order to be able to write T (ε)( f )(x) and

T (N)( f )(x) as convergent integrals for almost every x.

At this point we take absolute values, average over |z1− x| ≤ ε2

and |z2− x| ≤ N2

,

and we apply Holder’s inequality in two terms. We obtain the estimate

|T (ε ,N)( f )(x)|

≤ 1

vn

(2

ε

)n ∫

|z1−x|≤ ε2

∫

|x−y|≥ε|K(x− y)−K(z1− y)| | f (y)|dydz1

+1

vn

(2

ε

)n ∫

|z1−x|≤ ε2

|T ( f )(z1)|dz1

+

(1

vn

(2

ε

)n ∫

|z1−x|≤ ε2

|T ( f χ|x−·|<ε)(z1)|p dz1

) 1p

+1

vn

(2

N

)n ∫

|z2−x|≤ N2

∫

|x−y|≥N|K(x− y)−K(z2− y)| | f (y)|dydz2

+1

vn

(2

N

)n ∫

|z2−x|≤ N2

|T ( f )(z2)|dz2

+

(1

vn

(2

N

)n ∫

|z2−x|≤ N2

|T ( f χ|x−·|<N)(z2)|p dz2

) 1p

,

where vn is the volume of the unit ball in Rn. Applying condition (5.4.2) and estimate

(5.4.15), we obtain for f in Lp(Rn)∩L∞(Rn) with compact support that

|T (ε ,N)( f )(x)|

≤ 1

vn

(2

ε

)n ∫

|z1−x|≤ ε2

|T ( f )(z1)|dz1 +1

vn

(2

N

)n ∫

|z2−x|≤ N2

|T ( f )(z2)|dz2

+ cn

( 3

∑j=1

A j

)max(p,(p−1)−1)

(1

vn

(2

ε

)n ∫

|z1−x|≤ε| f (z1)|p dz1

) 1p

+ cn

( 3

∑j=1

A j

)max(p,(p−1)−1)

(1

vn

(2

N

)n ∫

|z2−x|≤N| f (z2)|p dz2

) 1p

+2A2

∥∥ f∥∥

L∞.

We now use density to remove the compact support condition on f and obtain the

last displayed estimate for all functions f in Lp(Rn)∩L∞(Rn). Taking the supremum

over all 0 < ε < N and over all N > 0, we deduce that for all f in Lp(Rn)∩L∞(Rn)we have the estimate

T (∗∗)( f )(x)≤ 2A2

∥∥ f∥∥

L∞+Sp( f )(x), (5.4.16)

where Sp is the sublinear operator defined by

Sp( f )(x) = 2M(T ( f ))(x)+3n+1cn

( 3

∑j=1

A j

)max(p,(p−1)−1)(M(| f |p)(x))

1p ,

and M is the Hardy–Littlewood maximal operator.

Recalling that M maps L1 to L1,∞ with bound at most 3n and also Lp to Lp,∞ with

bound at most 2 · 3n/p for 1 < p < ∞ (Exercise 2.1.4), we conclude that Sp maps

Lp(Rn) to Lp,∞(Rn) with norm at most

∥∥Sp

∥∥Lp→Lp,∞ ≤ cn(A1 +A2 +A3)max(p,(p−1)−1) , (5.4.17)

where cn is another dimensional constant.

Now write f = fα + f α , where

fα = f χ| f |≤α/(16A2) and f α = f χ| f |>α/(16A2).

The function fα is in L∞∩Lp and f α is in L1∩Lp. Moreover, we see that

∥∥ f α∥∥

L1 ≤ (16A2/α)p−1

∥∥ f∥∥p

Lp . (5.4.18)

Apply the Calderon–Zygmund decomposition (Theorem 5.3.1) to the function f α at

height αγ to write f α = gα + bα , where gα is the good function and bα is the bad

function of this decomposition. Using (5.3.1), we obtain

∥∥gα∥∥

Lp ≤ 2n/p′(αγ)1/p′∥∥ f α∥∥1/p

L1 ≤ 2(n+4)/p′(A2γ)1/p′∥∥ f

∥∥Lp . (5.4.19)

We now use (5.4.16) to get

|x ∈ Rn : T (∗∗)( f )(x)> α| ≤ b1 +b2 +b3 , (5.4.20)

where

b1 = |x ∈ Rn : 2A2

∥∥ fα∥∥

L∞+Sp( fα)(x)> α/4| ,

b2 = |x ∈ Rn : 2A2

∥∥gα∥∥

L∞+Sp(g

α)(x)> α/4| ,b3 = |x ∈ Rn : T (∗∗)(bα)(x)> α/2| .

Observe that 2A2‖ fα‖L∞ ≤ α/8. Selecting γ = 2−n−5(A1 +A2)−1 and using prop-

erty (2) in Theorem 5.3.1, we obtain

2A2

∥∥gα∥∥

L∞≤ A22n+1αγ ≤ α2−4 <

α

8

and therefore

b1 ≤ |x ∈ Rn : Sp( fα)(x)> α/8| ,b2 ≤ |x ∈ Rn : Sp(g

α)(x)> α/8| . (5.4.21)

Since γ ≤ (2n+5A1)−1, it follows from (5.3.27) that

b3 ≤∣∣∣⋃

j

Q∗j

∣∣∣+2n+8A2

∥∥ f α∥∥

L1

α≤

((5√

n)n

γ+2n+8A2

)∥∥ f α∥∥

L1

α,

and using (5.4.18), we obtain

b3 ≤Cn(A1 +A2)pα−p

∥∥ f∥∥p

Lp .

Using Chebyshev’s inequality in (5.4.21) and (5.4.17), we finally obtain that

b1 +b2 ≤ (8/α)p (cn)p(A1+A2+A3)

p max(p,(p−1)−1)p(∥∥ f

∥∥p

Lp +∥∥gα

∥∥p

Lp

).

Combining the estimates for b1,b2, and b3 and using (5.4.19), we deduce

∥∥T (∗∗)( f )∥∥

Lp,∞ ≤Cn(A1 +A2 +A3)max(p,(p−1)−1)∥∥ f

∥∥Lp(Rn)

. (5.4.22)

Finally, we need to obtain a similar estimate to (5.4.22), in which the weak Lp norm

on the left is replaced by the Lp norm. This is a consequence of Theorem 1.3.2 via

interpolation between the estimates Lp+1

2 → Lp+1

2 ,∞ and L2p → L2p,∞ for 2 < p < ∞and between the estimates L2p → L2p,∞ and L1 → L1,∞ for 1 < p < 2. The latter

estimate follows from Theorem 5.3.5. See also Corollary 5.3.7.

Exercises

5.4.1. Let T be a convolution operator that is L2 bounded. Suppose that a given

function f0 ∈ L1(Rn)∩L2(Rn) has vanishing integral and that T ( f0) is integrable.

Prove that T ( f0) also has vanishing integral.

5.4.2. Let K satisfy (5.4.1), (5.4.2), and (5.4.3) and let W ∈S ′ be an extension of

K on Rn. Let f be a compactly supported C 1 function on Rn with mean value zero.

Prove that the function f ∗W is in L1(Rn).

5.4.3. Suppose K is a function on Rn \0 that satisfies (5.4.1), (5.4.2), and (5.4.3).

Let K(ε ,N)(x) = K(x)χε<|x|<N for 0< ε < N <∞ and let T (ε ,N) be the operator given

by convolution with K(ε ,N). Let 1 < p < ∞ and f ∈ Lp(Rn). Prove that for some

sequence ε j ↓ 0, T (ε j ,N)( f ) converges almost everywhere as j→ ∞ and N → ∞.[Hint: Use Theorems 5.4.5 and 2.1.14.

]

5.4.4. (a) Prove that for all x,y ∈ Rn that satisfy 0 = x = y we have

∣∣∣∣x− y

|x− y| −x

|x|

∣∣∣∣≤ 2|y||x| .

(b) Let Ω be an integrable function with mean value zero on the sphere Sn−1. Sup-

pose that Ω satisfies a Lipschitz (Holder) condition of order 0 < α < 1 on Sn−1.

This means that

|Ω(θ1)−Ω(θ2)| ≤ B0|θ1−θ2|α

for all θ1,θ2 ∈ Sn−1. Prove that K(x) = Ω(x/|x|)/|x|n satisfies Hormander’s condi-

tion (5.4.3) with constant at most a multiple of B0 +‖Ω‖L∞ .

5.4.5. Let Ω be an L1 function on Sn−1 with mean value zero.

(a) Let ω∞(t) = sup|Ω(θ1)−Ω(θ2)| : θ1,θ2 ∈ Sn−1, |θ1−θ2| ≤ t and suppose

that the following Dini condition holds:

∫ 1

0ω∞(t)

dt

t< ∞ .

Prove that the function K(x)=Ω(x/|x|)|x|−n satisfies Hormander’s condition.

(b) (A. Calderon and A. Zygmund) For A ∈ O(n), let

∥∥A∥∥= sup|θ −A(θ)| : θ ∈ Sn−1 .

5.5 Vector-Valued Inequalities 385

Suppose that Ω satisfies the more general Dini-type condition

∫ 1

0ω1(t)

dt

t< ∞ ,

where

ω1(t) = supA∈O(n)‖A‖≤t

∫

Sn−1|Ω(A(θ))−Ω(θ)|dθ .

Prove the same conclusion as in part (a).[Hint: Part (b): Use the result in part (a) of Exercise 5.4.4 and switch to polar

coordinates.]

5.5 Vector-Valued Inequalities

Certain nonlinear expressions that appear in Fourier analysis, such as maximal func-

tions and square functions, can be viewed as linear quantities taking values in some

Banach space. This point of view provides the motivation for a systematic study of

Banach-valued operators. Let us illustrate this line of thinking via an example. Let

T be a linear operator acting on Lp of some measure space (X ,µ) and taking values

in the set of measurable functions of another measure space (Y,ν). The seemingly

nonlinear inequality

∥∥∥(∑

j

|T ( f j)|2)1

2∥∥∥

Lp≤Cp

∥∥∥(∑

j

| f j|2)1

2∥∥∥

Lp(5.5.1)

can be transformed to a linear one with only a slight change of view. Let us denote

by Lp(X , ℓ2) the Banach space of all sequences f j j of measurable functions on X

that satisfy

∥∥ f j j

∥∥Lp(X ,ℓ2)

=

(∫

X

(∑

j

| f j|2)p

2dµ

)1p

< ∞ . (5.5.2)

Define a linear operator acting on such sequences by setting

T ( f j j) = T ( f j) j. (5.5.3)

Then (5.5.1) is equivalent to the inequality

∥∥T ( f j j)∥∥

Lp(Y,ℓ2)≤Cp

∥∥ f j j

∥∥Lp(X ,ℓ2)

, (5.5.4)

in which T is thought of as a linear operator acting on the Lp space of ℓ2-valued

functions on X . This is the basic idea of vector-valued inequalities. A nonlinear

inequality such as (5.5.1) can be viewed as a linear norm estimate for an operator

acting and taking values in suitable Banach spaces.

5.5.1 ℓ2-Valued Extensions of Linear Operators

The following result is classical and fundamental in the subject of vector-valued

inequalities.

Theorem 5.5.1. Let 0 < p,q < ∞ and let (X ,µ) and (Y,ν) be two σ -finite measure

spaces. The following are valid:

(a) Suppose that T is a bounded linear operator from Lp(X) to Lq(Y ) with norm N.

Then T has an ℓ2-valued extension, that is, for all complex-valued functions f j in

Lp(X) we have

∥∥∥(∑

j

|T ( f j)|2)1

2∥∥∥

Lq(Y )≤Cp,q N

∥∥∥(∑

j

| f j|2)1

2∥∥∥

Lp(X)(5.5.5)

for some constant Cp,q that satisfies Cp,q = 1 if p ≤ q. Moreover, if T maps real-

valued Lp functions to real-valued Lq functions with norm Nreal , then (5.5.5) holds

for real-valued functions f j with Nreal in place of N.

(b) Suppose that T is a bounded linear operator from Lp(X) to Lq,∞(Y ) with norm

M. Then T has an ℓ2-valued extension, that is,

∥∥∥(∑

j

|T ( f j)|2)1

2∥∥∥

Lq,∞(Y )≤ Dp,q M

∥∥∥(∑

j

| f j|2)1

2∥∥∥

Lp(X)(5.5.6)

for some constant Dp,q that depends only on p and q. Moreover, if T maps real-

valued Lp functions to real-valued Lq,∞ functions with norm Mreal , then (5.5.5) holds

for real-valued functions f j with Mreal in place of M.

To prove this theorem, we need the following identities.

Lemma 5.5.2. For any 0 < r < ∞, define constants

Ar =

(Γ ( r+1

2)

πr+1

2

)1r

and Br =

(Γ ( r

2+1)

πr2

)1r

. (5.5.7)

Then for any λ1,λ2, . . . ,λn ∈ R we have

(∫

Rn|λ1x1 + · · ·+λnxn|re−π|x|

2

dx

)1r

= Ar (λ21 + · · ·+λ 2

n )12 , (5.5.8)

where dx = dx1 · · ·dxn. Also for all w1,w2, . . . ,wn ∈ C we have

(∫

Cn|w1z1 + · · ·+wnzn|re−π|z|

2

dz

)1r

= Br(|w1|2 + · · ·+ |wn|2)12 , (5.5.9)

where dz = dz1 · · ·dzn = dx1dy1 · · ·dxndyn if z j = x j + iy j.


Proof. Dividing both sides of (5.5.8) by (λ 21 + · · ·+λ 2

n )12 , we reduce things to the

situation in which λ 21 + · · ·+ λ 2

n = 1. Let e1 = (1,0, . . . ,0)t be the standard basis

column unit vector on Rn and find an orthogonal n×n matrix A ∈O(n) (orthogonal

means a real matrix satisfying At = A−1) such that A−1e1 = (λ1, . . . ,λn)t . Then the

first coordinate of Ax is

(Ax)1 = Ax · e1 = x ·Ate1 = x ·A−1e1 = λ1x1 + · · ·+λnxn .

Now change variables y = Ax in the integral in (5.5.8) and use the fact that |Ax|= |x|to obtain

(∫

Rn|λ1x1 + · · ·+λnxn|re−π|x|

2

dx

)1r

=

(∫

Rn|y1|re−π|y|

2

dy

)1r

=

(2

∫ ∞

0tre−πt2

dt

)1r

=

(∫ ∞

0s

r−12 e−πsds

)1r

=

(Γ ( r+1

2)

πr+1

2

)1r

= Ar ,

which proves (5.5.8). We used the fact that∫

R e−π|x|2dx = 1.

The proof of (5.5.9) is almost identical. We normalize by assuming that

|w1|2 + · · ·+ |wn|2 = 1 ,

and we let ε1 be the column vector of Cn having 1 in the first entry and zero

elsewhere. We find a unitary n× n matrix A such that A −1ε1 = (w1, . . . ,wn)t .

Unitary means A −1 = A ∗, where A ∗ is the conjugate transpose matrix of A ,

i.e., the matrix whose entries are the complex conjugates of A t and that satis-

fies u ·A v = A ∗u · v for all u,v ∈ Cn. Then (A z)1 = w1z1 + · · ·+wnzn and also

|A z| = |z|; therefore, changing variables ζ = A z in the integral in (5.5.9), we can

rewrite that integral as

(∫

Cn|ζ1|re−π|ζ |

2

dζ

)1r

=

(∫

C|ζ1|re−π|ζ1|2dζ1

)1r

=

(2π

∫ ∞

0tre−πt2

t dt

)1r

=

(π∫ ∞

0s

r2 e−πsds

)1r

= (NB−1q )q

∫

X

(∫

Cn|z1 f1 + · · ·+ zn fn|qe−π|z|

2

dz

)pq

dµ

qp

= (NB−1q )q

∫

X(Bq)

p( n

∑j=1

| f j|2) p

2dµ

qp

= Nq∥∥∥( n

∑j=1

| f j|2)1

2∥∥∥

q

Lp(X).

Note that we made use of Minkowski’s integral inequality (Exercise 1.1.6) in the last

inequality. Letting n→ ∞ proves the required conclusion with Cp,q = 1 if q≥ p.

If T happens to map real-valued functions to real-valued functions, then we adapt

the preceding argument by taking f j to be real-valued functions, we replace Bp by

Ap, Bq by Aq, N by Nreal and we use identity (5.5.8) instead of (5.5.9).

Part (b): Inequality (5.5.6) will be a consequence of (5.5.5) and of the following

result of Exercise 1.1.12 which holds when Y is σ -finite:

‖g‖Lq,∞ ≤ sup0<ν(E)<∞

ν(E)1q− 1

r

(∫

E|g|r dν

)1r

≤( q

q− r

)1r ‖g‖Lq,∞ , (5.5.10)

where 0 < r < q and the supremum is taken over all subsets E of Y of finite measure.

Using (5.5.10), we obtain

∥∥∥(∑

j

|T ( f j)|2)1

2∥∥∥

Lq,∞(Y )

≤ sup0<ν(E)<∞

ν(E)1q− 1

r

(∫

E

(∑

j

|T ( f j)|2)r

2dν

)1r

= sup0<ν(E)<∞

ν(E)1q− 1

r

(∫

Y

(∑

j

|χE T ( f j)|2)r

2dν

)1r

≤ sup0<ν(E)<∞

ν(E)1q− 1

r∥∥TE

∥∥Lp→LrCp,r

(∫

X

(∑

j

| f j|2)p

2dµ

)1p

, (5.5.11)

where TE is defined by TE( f ) = χE T ( f ). Since for any function f in Lp(X) we have

ν(E)1q− 1

r∥∥TE( f )

∥∥Lr ≤

( q

q− r

)1r ∥∥T ( f )

∥∥Lq,∞ ≤

( q

q− r

)1rM∥∥ f

∥∥Lp ,

it follows that for any measurable set E of finite measure the estimate

ν(E)1q− 1

r∥∥TE

∥∥Lp→Lr ≤

( q

q− r

)1rM (5.5.12)

is valid. Inserting (5.5.12) in (5.5.11), we obtain (5.5.6) with Dp,q =Cp,r

(q

q−r

)1r and

0 < r < q. Recall that Cp,r = 1 if r ≥ p and Cp,r = BpB−1r if r < p.

If T happens to map real-valued functions to real-valued functions, then we take

f j to be real-valued and we replace M by Mreal , Br by Ar, and Bq by Aq in the

preceding argument.

5.5.2 Applications and ℓr-Valued Extensions of Linear Operators

Here is an application of Theorem 5.5.1:

Example 5.5.3. On the real line consider the intervals I j = [b j,∞) for j ∈ Z. Let Tj

be the operator given by multiplication on the Fourier transform by the characteristic

function of I j. Then we have the following two inequalities:

∥∥∥(∑j∈Z

|Tj( f j)|2)1

2∥∥∥

Lp(R)≤Cp

∥∥∥(∑j∈Z

| f j|2)1

2∥∥∥

Lp(R), (5.5.13)

∥∥∥(∑j∈Z

|Tj( f j)|2)1

2∥∥∥

L1,∞(R)≤C

∥∥∥(∑j∈Z

| f j|2)1

2∥∥∥

L1(R), (5.5.14)

for 1 < p <∞. To prove these, first observe that the operator T = 12(I+ iH) is given

on the Fourier transform by multiplication by the characteristic function of the half-

axis [0,∞) [precisely, the Fourier multiplier of T is equal to 1 on the set (0,∞)and 1/2 at the origin; this function is almost everywhere equal to the characteristic

function of the half-axis [0,∞)]. Moreover, each Tj is given by

Tj( f )(x) = e2πib jxT (e−2πib j( ·) f )(x)

and thus with g j(x) = e−2πib jx f (x), (5.5.13) and (5.5.14) can be written respectively

as

∥∥∥(∑j∈Z

|T (g j)|2)1

2∥∥∥

Lp(R)≤Cp

∥∥∥(∑j∈Z

|g j|2)1

2∥∥∥

Lp(R),

∥∥∥(∑j∈Z

|T (g j)|2)1

2∥∥∥

L1,∞(R)≤C

∥∥∥(∑j∈Z

|g j|2)1

2∥∥∥

L1(R).

Theorem 5.5.1 gives that both of the previous estimates are valid by in view of the

boundedness of T = 12(I + iH) from Lp to Lp and from L1 → L1,∞. For a slight

generalization and an extension to higher dimensions, see Exercise 5.6.1.

We have now seen that bounded operators from Lp to Lq (or to Lq,∞) always

admit ℓ2-valued extensions. It is natural to ask whether they also admit ℓr-valued

extensions for some r = 2. For some values of r we may answer this question. Here

is a straightforward corollary of Theorem 5.5.1.

Corollary 5.5.4. Let (X ,µ) and (Y,ν) be σ -finite measure spaces. Suppose that T

is a linear bounded operator from Lp(X) to Lp(Y ) with norm A for some 1 < p <∞.

Let r be a number between p and 2. Then we have

∥∥∥(∑

j

|T ( f j)|r)1

r∥∥∥

Lp(Y )≤ A

∥∥∥(∑

j

| f j|r)1

r∥∥∥

Lp(X). (5.5.15)

Proof. Using Exercise 5.5.2 we interpolate between the trivial bound Lp(X , ℓp)→Lp(Y, ℓp) and the bound Lp(X , ℓ2)→ Lp(Y, ℓ2), which follows from Theorem 5.5.1.

We obtain the bound Lp(X , ℓr)→ Lp(Y, ℓr) since r lies between p and 2.

Example 5.5.5. The result of Corollary 5.5.4 may fail if r does not lie in the interval

with endpoints p and 2. Let us take, for example, 1 < p < 2 and consider an r < p.

Take X = Y = R and define a linear operator T by setting

T ( f )(x) = f (x)χ[0,1](x).

Then T is Lp bounded, since ‖T ( f )‖Lp ≤‖T ( f )‖Lp′ ≤‖ f‖Lp . Now take f j = χ[ j−1, j]

for j = 1, . . . ,N. A simple calculation gives

( N

∑j=1

|T ( f j)(x)|r)1

r= N

1r

∣∣∣∣e−2πix−1

−2πixχ[0,1](x)

∣∣∣∣ ,

while ( N

∑j=1

| f j|r)1

r= χ[0,N] .

It follows that N1/r ≤CN1/p for all N > 1, and hence (5.5.15) cannot hold if p > r.

We have now seen that ℓr-valued extensions for r = 2 may fail in general. But do

they fail for some specific operators of interest in Fourier analysis? For instance, is

the inequality ∥∥∥(∑j∈Z

|H( f j)|r)1

r∥∥∥

Lp≤Cp,r

∥∥∥(∑j∈Z

| f j|r)1

r∥∥∥

Lp(5.5.16)

true for the Hilbert transform H whenever 1 < p,r <∞? The answer to this question

is affirmative. Inequality (5.5.16) is indeed valid and was first proved using complex

function theory. In the next section we plan to study inequalities such as (5.5.16)

for general singular integrals using the Calderon–Zygmund theory of the previous

section applied to the context of Banach-valued functions.

5.5.3 General Banach-Valued Extensions

Let B be a Banach space over the field of complex numbers with norm ‖ ‖B , and let

B∗ be its dual (with norm ‖ ‖B∗ ). A function F defined on a measure space (X ,µ)

and taking values in B is called B-measurable if there exists a measurable subset

X0 of X such that µ(X \X0) = 0, F [X0] is contained in some separable subspace B0

of B, and for every u∗ ∈B∗ the complex-valued map

x →⟨u∗,F(x)

⟩

is measurable. A consequence of this definition is that the positive function x →‖F(x)‖B on X is measurable; to see this, use the relevant result in [382, p. 131].

For 0 < p≤∞ we denote by Lp(X) the space Lp(X ,C). Let Lp(X)⊗B be the set

of all finite linear combinations of elements of B with coefficients in Lp(X), that is,

elements of the form

F = f1u1 + · · ·+ fmum, (5.5.17)

where f j ∈ Lp(X), u j ∈B, and m ∈ Z+. We define Lp(X ,B) to be the space of all

B-measurable functions F on X satisfying

(∫

X

∥∥F(x)∥∥p

Bdµ(x)

)1p

< ∞ , (5.5.18)

with the obvious modification when p=∞. Similarly define Lp,∞(X ,B) as the space

of all B-measurable functions F on X satisfying

∥∥∥∥∥F( ·)

∥∥B

∥∥∥Lp,∞(X)

< ∞ . (5.5.19)

Then Lp(X ,B) (respectively, Lp,∞(X ,B)) is called the Lp (respectively, Lp,∞) space

of functions on X with values in B. The quantity in (5.5.18) (respectively, in (5.5.19))

is the norm of F in Lp(X ,B) (respectively, in Lp,∞(X ,B)).

Proposition 5.5.6. Let B be a Banach space and (X ,µ) a σ -finite measure space.

(a) The set ∑mj=1 χE j

u j : u j ∈B, E j X are pairwise disjoint and µ(E j)< ∞ is

dense in Lp(X ,B) whenever 0 < p < ∞.

(b) The set ∑∞j=0 χE ju j : u j ∈B, E j X are pairwise disjoint and X = ∪∞j=0E j is

dense in L∞(X ,B).(c) The space C ∞

0 ⊗B of functions of the form ∑mj=1ϕ j u j, where u j ∈B, ϕ j are in

C ∞0 (Rn), is dense in Lp(Rn,B) for 1≤ p < ∞.

Proof. If F ∈ Lp(X ,B) for 0 < p ≤ ∞, then F is B-measurable; thus there exists

X0 X satisfying µ(X \ X0) = 0 and F [X0] B0, where B0 is some separable

subspace of B. Choose a countable dense sequence u j∞j=1 of B0.

(a) First assume that p < ∞. Since X is σ -finite, for any ε > 0, there exists a

measurable subset X1 of X0 with µ(X1)< ∞ such that

∫

X\X1

∥∥F(x)∥∥p

Bdµ <

ε p

3.

Setting

B(u j,ε) =

u ∈B0 : ‖u−u j‖B < ε(3µ(X1))− 1

p,

we have B0 ⋃∞

j=1 B(u j,ε). Let A1 = B(u1,ε) and A j = B(u j,ε)\ (⋃ j−1

i=1 B(ui,ε))for j ≥ 2. It is easily seen that A j∞j=1 are pairwise disjoint and

⋃∞j=1 A j =⋃∞

j=1 B(u j,ε). Set E j = F−1[A j]∩ X1. Then X1 =⋃∞

j=1 E j and E j∞j=1 are pair-

wise disjoint. Since µ(X1) = ∑∞j=1 µ(E j) < ∞, it follows that µ(E j) < ∞ and also

that for some m ∈ Z+,

∫⋃∞

j=m+1 E j

‖F(x)‖p

Bdµ <

ε p

3. (5.5.20)

Moreover, one can easily verify that ∑mj=1 χE j

u j is B-measurable. Notice that

‖F(x)− u j‖B < ε(3µ(X1))−1/p for any x ∈ E j and j ∈ 1, . . . ,m. This fact com-

bined with (5.5.20) and the mutual disjointness of E jmj=1 yields that

∫

X

∥∥∥∥F(x)−m

∑j=1

χE j(x)u j

∥∥∥∥p

B

dµ =∫

X\X1

∥∥F(x)∥∥p

Bdµ+

∫

∪∞j=m+1E j

∥∥F(x)∥∥p

Bdµ

+∫⋃m

j=1 E j

∥∥∥∥m

∑j=1

χE j(x)[F(x)−u j]

∥∥∥∥p

B

dµ

<ε p

3+ε p

3+ε p

3= ε p .

(b) Now consider the case p =∞. Obviously we have B0 ⋃∞

j=1 B(u j,ε), where

B(u j,ε) = u ∈B0 : ‖u− u j‖B < ε. Let A1 = B(u1,ε) and for j ≥ 2 define sets

A j = B(u j,ε)\ (⋃ j−1

i=1 B(ui,ε)). Let E j = F−1[A j] for j ≥ 1 and E0 = X \ (⋃∞j=1 E j).

Then µ(E0) = 0. As in the proof of the case p < ∞, we have that E j∞j=0 are

pairwise disjoint and X0 ⋃∞

j=0 E j. Pick u0 = 0. Notice that ∑∞j=0 χE ju j is B-

measurable. Since ‖F(x)−u j‖B < ε for any x ∈ E j and j ≥ 0, we have

∥∥∥∥F−∞

∑j=0

χE ju j

∥∥∥∥L∞(X ,B)

=

∥∥∥∥∞

∑j=0

χE j(F−u j)

∥∥∥∥L∞(X ,B)

< ε ,

which completes the proof in the case p = ∞.

(c) For the last assertion, we fix a smooth function with supported in the unit ball

of Rn with integral one. Let ϕδ (x) = δ−nϕ(x/δ ) for x ∈ Rn and δ > 0. Given a

function ∑mj=1 χE j

u j as in part (a) approximating a given f in Lp⊗B, we consider

the function ∑mj=1(χE j

∗ϕδ )u j which lies in C ∞0 ⊗B. Since ‖χE j

∗ϕδ −χE j‖Lp → 0

as δ → 0 when 1 ≤ p < ∞, ∑mj=1(χE j

∗ϕδ )u j tends to ∑mj=1 χE j

u j in Lp⊗B as

δ → 0, and the conclusion follows.

Let (X ,µ) be a measure space. If F is an element of L1(X)⊗B given as in

(5.5.17), we define its integral (which is an element of B) by setting

∫

XF(x)dµ(x) =

m

∑j=1

(∫

Xf j(x)dµ(x)

)u j.

Observe that for every F ∈ L1(X)⊗B we have

∥∥∥∫

XF(x)dµ(x)

∥∥∥B

= sup‖u∗‖B∗≤1

∣∣∣∣⟨

u∗,m

∑j=1

(∫

Xf j dµ

)u j

⟩∣∣∣∣

= sup‖u∗‖B∗≤1

∣∣∣∣∫

X

⟨u∗,

m

∑j=1

f ju j

⟩dµ

∣∣∣∣

≤∫

Xsup

‖u∗‖B∗≤1

∣∣⟨u∗,m

∑j=1

f ju j

⟩∣∣dµ

=∥∥F

∥∥L1(X ,B)

.

Thus the linear operator

F → IF =∫

XF(x)dµ(x)

is bounded from L1(X)⊗B into B. Since every element of L1(X ,B) is a (norm)

limit (Proposition 5.5.6 (c)) of a sequence of elements in L1(X)⊗B, by continuity,

the operator F → IF has a unique extension on L1(X ,B) that we call the Bochner

integral of F and denote by ∫

XF(x)dµ(x) .

L1(X ,B) is called the space of all Bochner integrable functions from X to B. Since

the Bochner integral is an extension of IF , for each F ∈ L1(X ,B) we have

∥∥∥∫

XF(x)dx

∥∥∥B≤

∫

X

∥∥F(x)∥∥

Bdx .

Consequently, measurable functions F with∫

X ‖F(x)‖B dx < ∞ are Bochner inte-

grable over X . It is not difficult to show that the Bochner integral of F is the only

element of B that satisfies

⟨u∗,

∫

XF(x)dµ(x)

⟩=

∫

X

⟨u∗,F(x)

⟩dµ(x) (5.5.21)

for all u∗ ∈B∗. The next result concerns duality in this context when X = Rn.

Proposition 5.5.7. Let B be a Banach space and 1≤ p≤ ∞.

(a) For any F ∈ Lp(Rn,B) we have

‖F‖Lp(Rn,B) = sup‖G‖

Lp′ (Rn,B∗)≤1

∣∣∣∣∫

Rn〈G(x),F(x)〉dx

∣∣∣∣ .

Consequently, Lp(Rn,B) isometrically embeds in (Lp′(Rn,B∗))∗.(b) for any G ∈ Lp′(Rn,B∗) one has

‖G‖Lp′ (Rn,B∗) = sup

‖F‖Lp(Rn,B)≤1

∣∣∣∣∫

Rn〈G(x),F(x)〉dx

∣∣∣∣

and thus Lp′(Rn,B∗) isometrically embeds in (Lp(Rn,B))∗.

Proof. Holder’s inequality yields that the right-hand side of (a) is controlled by its

left-hand side. It remains to establish the reverse inequality.

For F ∈ Lp(Rn,B) and ε > 0, by Proposition 5.5.6, there is Fε(x)=∑mj=1 χE j

(x)u j

with m ∈ Z+ or m = ∞ (when p = ∞) such that ‖Fε −F‖Lp(Rn,B) < ε/2, where

E jmj=1 are pairwise disjoint subsets of Rn and u j ∈B. Since Fε ∈ Lp(Rn,B), we

choose a nonnegative function h satisfying ‖h‖Lp′ (Rn)

≤ 1 such that

∥∥Fε∥∥

Lp(Rn,B)=

(∫

Rn

∥∥Fε(x)∥∥p

Bdx

) 1p

<∫

Rnh(x)

∥∥Fε(x)∥∥

Bdx+

ε

4. (5.5.22)

When 1≤ p < ∞, we can further choose h ∈ Lp′(Rn) to be a function with bounded

support, which ensures that it is integrable. For given u j ∈B, there exists u∗j ∈B∗

satisfying ‖u∗j‖B∗ = 1 and

‖u j‖B < 〈u∗j ,u j〉+ε

4(‖h‖L1(Rn)+1). (5.5.23)

Set G(x) =∑mj=1 h(x)χE j

(x)u∗j . Clearly G is B∗-measurable and ‖G‖Lp′ (Rn,B∗) ≤ 1.

It follows from (5.5.22) and (5.5.23) that

∫

Rn〈G(x),Fε(x)〉dx =

∫

Rnh(x)

m

∑j=1

χE j(x)〈u∗j ,u j〉dx

≥∫

Rnh(x)

m

∑j=1

(‖u j‖B−

ε

4(‖h‖L1(Rn)+1)

)χE j

(x)dx

≥∥∥Fε

∥∥Lp(Rn,B)

− ε

4− ε

4.

Using Holder’s inequality we have

∣∣∣∣∫

Rn〈G(x),Fε(x)−F(x)〉dx

∣∣∣∣≤ ‖G‖Lp′ (Rn,B∗)‖F−Fε‖Lp(Rn,B) <ε

2,

hence we obtain

∥∥Fε∥∥

Lp(Rn,B)≤ sup‖G‖

Lp′ (Rn,B∗)≤1

∣∣∣∣∫

Rn〈G(x),F(x)〉dx

∣∣∣∣+ ε .

Letting ε → 0 yields the desired inequality in part (a).

(b) The duality statement |〈G(x),F(x)〉| ≤ ‖G(x)‖B∗‖F(x)‖B together with

Holder’s inequality imply the ≥ inequality in part (b). We prove the ≤ inequal-

ity via an argument symmetric with that in case (a). For completeness we in-

clude the details. For a given G in Lp′(Rn,B) and ε > 0, by Proposition 5.5.6,

there is Gε(x) =∑mj=1 χE j

(x)u∗j with m ∈ Z+ or m = ∞ (when p = 1) such that

‖Gε −G‖Lp′ (Rn,B∗) < ε/2, where E jm

j=1 are pairwise disjoint subsets of Rn and

u∗j ∈B∗. Since Gε lies in Lp′(Rn,B∗), we choose a nonnegative function h satisfy-

ing ‖h‖Lp(Rn) ≤ 1 such that

∥∥Gε

∥∥Lp′ (Rn,B∗) =

(∫

Rn

∥∥Gε(x)∥∥p′

B∗ dx

) 1p′<

∫

Rnh(x)

∥∥Gε(x)∥∥

B∗ dx+ε

4. (5.5.24)

When 1 < p≤ ∞, we can further choose h ∈ Lp(Rn) to be a function with bounded

support, which ensures that it is integrable. For each u∗j ∈B, there exists u j ∈B

satisfying ‖u j‖B = 1 and

‖u∗j‖B∗ < 〈u∗j ,u j〉+ε

4(‖h‖L1(Rn)+1). (5.5.25)

Set F(x) = ∑mj=1 h(x)χE j

(x)u j. Clearly F is B-measurable and ‖F‖Lp(Rn,B) ≤ 1. It

follows from (5.5.24) and (5.5.25) that

∫

Rn〈Gε(x),F(x)〉dx =

∫

Rnh(x)

m

∑j=1

χE j(x)〈u∗j ,u j〉dx

≥∫

Rnh(x)

m

∑j=1

(‖u∗j‖B∗ − ε

4(‖h‖L1(Rn)+1)

)χE j

(x)dx

≥∥∥Gε

∥∥Lp′ (Rn,B∗)−

ε

2.

Hence, for any ε > 0, we have

∥∥Gε

∥∥Lp′ (Rn,B∗) ≤ sup

‖F‖Lp(Rn,B)≤1

∣∣∣∣∫

Rn〈G(x),F(x)〉dx

∣∣∣∣+ ε

which implies the reverse inequality in part (b) by letting ε → 0.

Definition 5.5.8. Let X ,Y be measure spaces. Let T be a linear operator that maps

Lp(X) to Lq(Y ) (respectively, Lp(X) to Lq,∞(Y )) for some 0 < p,q ≤ ∞. We define

another operator T acting on Lp(X)⊗B by setting

T( m

∑j=1

f ju j

)=

m

∑j=1

T ( f j)u j.

If T happens to have a bounded extension from Lp(X ,B) to Lq(Y,B) (respectively

from Lp(X ,B) to Lq,∞(Y,B)), then we say that T has a bounded B-valued exten-

sion. In this case we also denote by T the B-valued extension of T .

Example 5.5.9. Let B = ℓr for some 1 ≤ r < ∞. Then a measurable function

F : X → B is just a sequence f j j of measurable functions f j : X → C. The

space Lp(X , ℓr) consists of all measurable complex-valued sequences f j j on X

that satisfy∥∥ f j j

∥∥Lp(X ,ℓr)

=∥∥∥(∑

j

| f j|r)1

r∥∥∥

Lp(X)< ∞ .

The space Lp(X)⊗ ℓr is the set of all finite sums

m

∑j=1

(a j1,a j2,a j3, . . .)g j ,

where g j ∈ Lp(X) and (a j1,a j2,a j3, . . .) ∈ ℓr, j = 1, . . . ,m. This is certainly a sub-

space of Lp(X , ℓr). Now given ( f1, f2, . . .) ∈ Lp(X , ℓr), let Fm = e1 f1 + · · ·+ em fm,

where e j is the infinite sequence with zeros everywhere except at the jth entry, where

it has 1. Then Fm ∈ Lp(X)⊗ ℓr and approximates f in the norm of Lp(X , ℓr). This

shows the density of Lp(X)⊗ ℓr in Lp(X , ℓr).If T is a linear operator bounded from Lp(X) to Lq(Y ), then T is defined by

T ( f j j) = T ( f j) j.

According to Definition 5.5.8, T has a bounded ℓr-extension if and only if the in-

equality ∥∥∥(∑

j

|T ( f j)|r)1

r∥∥∥

Lq≤C

∥∥∥(∑

j

| f j|r)1

r∥∥∥

Lp

is valid.

A linear operator T acting on measurable functions is called positive if it satisfies

f ≥ 0 =⇒ T ( f )≥ 0. It is straightforward to verify that positive operators satisfy

f ≤ g =⇒ T ( f )≤ T (g) ,

|T ( f )| ≤ T (| f |) ,sup

j

|T ( f j)| ≤ T(

supj

| f j|),

(5.5.26)

for all f ,g, f j measurable functions. We have the following result regarding vector-

valued extensions of positive operators:

Proposition 5.5.10. Let 0 < p,q≤ ∞ and (X ,µ), (Y,ν) be two measure spaces. Let

T be a positive linear operator mapping Lp(X) to Lq(Y ) (respectively, to Lq,∞(Y ))with norm A. Let B be a Banach space. Then T has a B-valued extension T that

maps Lp(X ,B) to Lq(Y,B) (respectively, to Lq,∞(Y,B)) with the same norm.

Proof. Let us first understand this theorem when B = ℓr for 1 ≤ r ≤ ∞. The two

endpoint cases r = 1 and r =∞ can be checked easily using the properties in (5.5.26).

For instance, for r = 1 we have

∥∥∥∑j

|T ( f j)|∥∥∥

Lq≤

∥∥∥∑j

T (| f j|)∥∥∥

Lq=

∥∥∥T(∑

j

| f j|)∥∥∥

Lq≤ A

∥∥∥∑j

| f j|∥∥∥

Lp,

while for r = ∞ we have∥∥∥sup

j

|T ( f j)|∥∥∥

Lq≤

∥∥∥T (supj

| f j|)∥∥∥

Lq≤ A

∥∥supj

| f j|∥∥

Lp .

The required inequality for 1 < r < ∞,

∥∥∥(∑

j

|T ( f j)|r)1

r∥∥∥

Lq≤ A

∥∥∥(∑

j

| f j|r)1

r∥∥∥

Lp,

follows from the Riesz–Thorin interpolation theorem (see Exercise 5.5.2).

The result for a general Banach space B can be proved using the following in-

equality: ∥∥T (F)(x)∥∥

B≤ T

(‖F‖B

)(x), x ∈ X , (5.5.27)

by simply taking Lq norms. To prove (5.5.27), let us take F = ∑nj=1 f ju j. Then

∥∥T (F)(x)∥∥

B=

∥∥∥n

∑j=1

T ( f j)(x)u j

∥∥∥B

= sup‖u∗‖B∗≤1

∣∣∣⟨u∗,

n

∑j=1

T ( f j)(x)u j

⟩∣∣∣

= sup‖u∗‖B∗≤1

∣∣∣T( n

∑j=1

f j

⟨u∗,u j

⟩)(x)

∣∣∣

≤ T(

sup‖u∗‖B∗≤1

∣∣⟨u∗,n

∑j=1

f ju j

⟩∣∣)(x)

= T(∥∥

n

∑j=1

f ju j

∥∥B

)(x)

= T(‖F‖B

)(x) ,

where the inequality makes use of the fact that T is a positive operator.

Exercises

5.5.1. ([207]) Let (X ,µ) and (Y,ν) be σ -finite measure spaces and suppose that

0 < p1, p1 ≤ ∞, 1≤ q0,q1 ≤ ∞, and that p0 > p1. For 0 < θ < 1 define p,q by

1

p=

1−θp0

+θ

p1,

1

q=

1−θq0

+θ

q1.

Let B1 and B2 be Banach spaces and let T be a linear operator that maps

Lp0(X ,B1) to Lq0(Y,B2) with norm A0 and Lp1(X ,B1) to Lq1(Y,B2) with norm

A1. Show that T has an extension that maps Lp(X ,B1) to Lq(Y,B2) with norm at

most 9A1−θ0 Aθ1 , by following the steps below:

(a) Let i ∈ 0,1. If Fj, j = 1, . . . ,m are in Lpi(B1) with disjoint supports, using

max j∈1,...,m ‖T (Fj)(y)‖B2≤ 1

2m ∑εk=±1

∥∥∑mk=1 εk

T (Fk)(y)∥∥

B2for y ∈ Y show that

∥∥∥ maxj=1,...,m

‖T (Fj)‖B2

∥∥∥Lqi≤ Ai

∥∥∥m

∑j=1

Fj

∥∥∥Lpi (X ,B1)

.

(b) Assume that F lies in a dense subspace of B1, it satisfies ‖‖F‖B1‖Lp = 1, and

it takes only finitely many values. For a λ > 1 pick a large integer N such that

λ−N < ‖F(x)‖B1≤ λN for all x ∈ X such that ‖F(x)‖B1

= 0 and define Fj = FχΩ j,

where Ω j = x : 2 j < ‖F‖B1≤ 2 j+1. Let a = p

p1− p

p0. Prove the inequalities

∥∥∥∑j

λ− jaθFj

∥∥∥p0

Lp0 (X ,B1)≤ λ aθ p0 and

∥∥∥∑j

λ jaθFj

∥∥∥p1

Lp1 (X ,B1)≤ 1 .

(c) Define g0(y) = max j λ− jaθ‖T (Fj)(y)‖B2

, g1(y) = max j λja(1−θ)‖T (Fj)(y)‖B2

for y ∈ Y and show that

‖g0‖Lq0 (Y ) ≤ A0λaθ and ‖g1‖Lq1 (Y ) ≤ A1 .

(d) Prove that for all y ∈ Y we have

‖T (F)(y)‖B2≤∑

j

‖T (Fj)(y)‖B2≤ g0(y)

1−θg1(y)θ

(2+

1

λ aθ −1+

1

λ a(1−θ)−1

)

and conclude that ‖T (F)‖Lp(Y,B2) ≤ 9A1−θ0 Aθ1 by picking λ = (1+

√2)aθ .[

Hint: Part (d): Split the sum according to whether λ ja > g1(y)g0(y)

and λ ja ≤ g1(y)g0(y)

.]

5.5.2. Prove the following version of the Riesz–Thorin interpolation theorem. Let

(X ,µ) and (Y,ν) be σ -finite measure spaces. Let 1 < p0,q0, , p1,q1,r0,s0,r1,s1 <∞and 0 < θ < 1 satisfy

1−θp0

+θ

p1=

1

p,

1−θq0

+θ

q1=

1

q,

1−θr0

+θ

r1=

1

r,

1−θs0

+θ

s1=

1

s.

Suppose that T is a linear operator that maps Lp0(X) to Lq0(Y ) and Lp1(X) to

Lq1(Y ). Define a vector-valued operator T by setting T ( f j j) = T ( f j) j acting

on sequences of complex-valued functions defined on X . Suppose that T maps

Lp0(X , ℓr0(C)) to Lq0(Y, ℓs0(C)) with norm M0 and Lp1(X , ℓr1(C)) to Lq1(Y, ℓs1(C))with norm M1. Prove that T maps Lp(X , ℓr(C)) to Lq(Y, ℓs(C)) with norm at most

M1−θ0 Mθ

1 .[Hint: Use the idea of the proof of Theorem 1.3.4. Apply Lemma 1.3.5 to the

function

F(z) =m

∑k=1

n

∑j=1∑l∈Z

aP(z)k,l eiαk,l

‖ak,ll‖R(z)−P(z)ℓr

bQ(z)j,l eiβ j,l

‖bk,ll‖S(z)−Q(z)

ℓs′

∫

YT (χAk

)(y)χB j(y)dν(y)

where P(z) = pp0(1− z)+ p

p1z, Q(z) = q′

q′0(1− z)+ q′

q′1z , R(z) = r

r0(1− z)+ r

r1z, and

S(z) = s′s′0(1− z)+ s′

s′1z, ak,ll∈Z,b j,ll∈Z are finitely supported sequences of pos-

itive reals, αk,l ,βk,l ∈ R, Ak are pairwise disjoint subsets of X with finite measure,

and B j are pairwise disjoint subsets of Y with finite measure.]

5.5.3. Prove the following version of the Marcinkiewicz interpolation theorem. Let

(X ,µ) and (Y,ν) be σ -finite measure spaces and let 0 < p0 < p < p1 ≤ ∞ and

0 < θ < 1 satisfy1−θ

p0+θ

p1=

1

p.

Let B1,B2 be Banach spaces and T be defined on Lp0(X ,B1)+Lp1(X ,B1) such

that for every y ∈ Y and for all F,G ∈ Lp0(X ,B1)+Lp1(X ,B1) we have

∥∥T (F +G)(y)∥∥

B2≤

∥∥T (F)(y)∥∥

B2+

∥∥T (G)(y)∥∥

B2.

(a) Suppose that T maps Lp0(X ,B1) to Lp0,∞(Y,B2) with norm A0 and Lp1(X ,B1)to Lp1,∞(Y,B2) with norm A1. Show that T maps Lp(X ,B1) to Lp(Y,B2) with norm

at most 2(

pp−p0

+ pp1−p

) 1p A1−θ

0 Aθ1 .

(b) Let p0 = 1. If T is linear and maps L1(X ,B1) to L1,∞(Y,B2) with norm A0 and

Lp1(X ,B1) to Lp1(Y,B2) with norm A1, show that T maps Lp(X ,B1) to Lp(Y,B2)

with norm at most 72(

p−1)−1/p

A1−θ0 Aθ1 .[

Hint: Part (a): Copy the proof of Theorem 1.3.2. Part (b): Use Exercise 5.5.1.]

5.5.4. For all x ∈ Rn let K(x) be a bounded linear operator from B1 to B2 and

let Q⊗B1 the space of all finite linear combinations of elements of the form F =

∑mi=1 χEi

ui , where Ei are disjoint measurable subsets of Rn of finite measure, ui in

B1, and m ∈ Z+.

(a) Suppose that K satisfies

∫

Rn

∥∥K(x)∥∥

B1→B2dx =C1 < ∞ .

Prove that the operator

T (F)(x) =∫

Rn

K(x− y)(F(y))dy ,

initially defined on Q⊗B1, has an extension that maps Lp(Rn,B1) to Lp(Rn,B2)with norm at most C1 for 1≤ p < ∞.

(b) (Young’s inequality) Let 1 ≤ p,q,s ≤ ∞ be such that p < ∞, s > 1, 1/q+ 1 =1/s+1/p. Suppose that K satisfies

∥∥∥∥∥K( ·)

∥∥B1→B2

∥∥∥Ls(Rn)

=C2 < ∞ .

5.6 Vector-Valued Singular Integrals 401

Prove that the T defined in part (a) has an extension that maps Lp(Rn,B1) to

Lq(Rn,B2) with norm at most C2.

(c) (Young’s inequality for weak type spaces) Suppose that 1 < p,s < ∞, 1/q+1 =1/s+1/p, and that K satisfies

∥∥∥∥∥K( ·)

∥∥B1→B2

∥∥∥Ls,∞(Rn)

=C3 < ∞ .

Prove that the T defined in part (a) has an extension that maps Lp(Rn,B1) to

Lq(Rn,B2).(d) Prove the following (slight) generalization of the assertion in part (a). Suppose

that K satisfies ∫

Rn

∥∥K(x)(u)∥∥

B2dx≤C1‖u‖B1

for all u ∈B1. Then T has an extension that maps L1(Rn,B1) to L1(Rn,B2) with

norm at most C1.

5.5.5. Use the inequality for the Rademacher functions in Appendix C.2 instead of

Lemma 5.5.2 to prove part (a) of Theorem 5.5.1 in the special case p = q. Notice

that this approach does not yield a sharp constant.

5.5.6. Let 0< p = 2≤∞ and suppose that Tj are uniformly bounded linear operators

from Lp(R) to Lp(R). Show that the inequality

∥∥∥(∑j∈Z

|Tj( f j)|2)1

2

∥∥∥Lp≤Cp

∥∥∥(∑j∈Z

| f j|2)1

2

∥∥∥Lp

may fail.[Hint: Let Tj(g)(x) = g(x− j). When p> 2 take f j(x) = χ[− j,1− j] for j = 1,2, . . . ,N.

When p < 2 take f j = χ[0,1] for j = 1,2, . . . ,N.]

5.5.7. Suppose that T is a linear operator that takes real-valued functions to real-

valued functions. Use Theorem 5.5.1(a) with p = q to prove that

supf real-valued

f =0

∥∥T ( f )∥∥

Lp∥∥ f∥∥

Lp

= supf complex-valued

f =0

∥∥T ( f )∥∥

Lp∥∥ f∥∥

Lp

.

5.6 Vector-Valued Singular Integrals

We now discuss some results about vector-valued singular integrals. By this we

mean singular integral operators acting on functions defined on Rn and taking values

in Banach spaces.

5.6.1 Banach-Valued Singular Integral Operators

Suppose that B1 and B2 are Banach spaces. We denote by L(B1,B2) the space

of all bounded linear operators from B1 to B2. We consider a kernel K defined on

Rn \ 0 that takes values in L(B1,B2). In other words, for all x ∈ Rn \ 0, K(x)is a bounded linear operator from B1 to B2 with norm ‖K(x)‖B1→B2

. Thus for any

v ∈B1 and any x ∈ Rn \0 we have

∥∥K(x)(v)∥∥

B2≤ ‖K(x)‖B1→B2

‖v‖B1.

We assume that there is a constant A < ∞ such that the size condition holds

∥∥K(x)∥∥

B1→B2≤ A |x|−n , (5.6.1)

and also the regularity condition

supy∈Rn\0

∫

|x|≥2|y|

∥∥K(x− y)−K(x)∥∥

B1→B2dx≤ A < ∞ . (5.6.2)

Moreover, we assume that there is a sequence εk ↓ 0 as k→ ∞ and an element K0 of

L(B1,B2) such that

limk→∞

∥∥∥∥∫

εk≤|y|≤1

K(y)dy− K0

∥∥∥∥B1→B2

= 0 . (5.6.3)

Given these assumptions, we define an operator T on C ∞0 ⊗B1 as follows:

For functions fi ∈ C ∞0 (Rn) and ui ∈B1 we define

T( m

∑i=1

fiui

)(x) = lim

k→∞

∫

εk≤|y|K(y)

( m

∑i=1

fi(x− y)ui

)dy (5.6.4)

=m

∑i=1

∫

|y|≤1( fi(x− y)− fi(x))K(y)(ui)dy+

m

∑i=1

fi(x)K0(ui)

+

∫

|y|>1

m

∑i=1

fi(x− y)K(y)(ui)dy . (5.6.5)

Notice that for each i ∈ 1, . . . ,m we have

∫

|y|≤1| fi(x− y)− fi(x)|‖K(y)(ui)‖B2

dy≤ ‖∇ fi‖L∞‖ui‖B1

∫

|y|≤1|y|‖K(y)‖B1→B2

dy

and this is a finite integral in view of (5.6.1). Thus the function

( fi(x− y)− fi(x))K(y)(ui)

is B2-integrable and the expression

∫

|y|≤1( fi(x− y)− fi(x))‖K(y)(ui)‖B2

dy

is a well-defined element of B2. Also the integral in (5.6.5) is over the compact set

1≤ |y| ≤ |x|+M, where the ball B(0,M) contains the supports of all fi, and thus it

also converges in B2, using (5.6.1).

The following vector-valued extension of Theorem 5.3.3 is the main result of this

section.

Theorem 5.6.1. Let B1 and B2 be Banach spaces. Suppose that K(x) satisfies

(5.6.1), (5.6.2), and (5.6.3) for some A > 0 and K0 ∈ L(B1,B2). Let T be the

operator associated with K as in (5.6.4). Assume that T is a bounded linear op-

erator from Lr(Rn,B1) to Lr(Rn,B2) with norm B⋆ for some 1 < r ≤ ∞. Then Thas well-defined extensions on Lp(Rn,B1) for all 1≤ p < ∞. Moreover, there exist

dimensional constants Cn and C′n such that

∥∥T (F)∥∥

L1,∞(Rn,B2)≤C′n(A+B⋆)

∥∥F∥∥

L1(Rn,B1)(5.6.6)

for all F in L1(Rn,B1) and

∥∥T (F)∥∥

Lp(Rn,B2)≤Cn max

(p,(p−1)−1

)(A+B⋆)

∥∥F∥∥

Lp(Rn,B1)(5.6.7)

whenever 1 < p < ∞ and F is in Lp(Rn,B1).

Proof. Although T is defined on the entire L1(Rn,B1)∩Lr(Rn,B1), it will be con-

venient to work with its restriction to a smaller dense subspace of L1(Rn,B1). We

make the observation that the space Q⊗B1 of all functions of the form ∑mi=1 χRi

ui,

where Ri are disjoint dyadic cubes and ui ∈B1, is dense in L1(Rn,B1). Indeed, by

Proposition 5.5.6 (b) it suffices to approximate a C ∞0 ⊗B1-valued function with a

Q⊗B1-valued function. But this is immediate since any function in C ∞0 (Rn) can

be approximated in L1(Rn) by finite linear combinations of characteristic functions

of disjoint dyadic cubes.

Case 1: r = ∞. We fix F = ∑mi=1 χRi

ui in Q⊗B1 and we notice that for each

x ∈ Rn we have ‖F(x)‖B1= ∑m

i=1 χRi(x)‖ui‖B1

, which is also a finite linear combi-

nation of characteristic functions of dyadic cubes. Apply the Calderon-Zygmund

decomposition to ‖F‖B1at height γα, where γ = 2−n−1B−1

⋆ as in the proof of

Theorem 6.3.1. We extract a finite collection of closed dyadic cubes Q j j satis-

fying ∑ j |Q j| ≤ (γα)−1‖F‖L1(Rn,B1)and we define the good function of the decom-

position

G(x) =

F(x) for x /∈ ∪ jQ j

|Q j|−1∫

Q jF(x)dx for x ∈ Q j.


Also define the bad function B(x) = F(x)−G(x). Then B(x) =∑ j B j(x), where each

B j is supported in Q j and has mean value zero over Q j. Moreover,

‖G‖L1(Rn,B1)≤ ‖F‖L1(Rn,B1)

(5.6.8)

‖G‖L∞(Rn,B1) ≤ 2nγα (5.6.9)

and ‖B j‖L1(Rn,B1)≤ 2n+1γα|Q j|, by an argument similar to that given in the proof

of Theorem 5.3.1. We only verify (5.6.9). On the cube Q j, G is equal to the constant

|Q j|−1∫

Q jF(x)dx, and this is bounded by 2nγα . For each x ∈ Rn \⋃ j Q j and for

each k = 0,1,2, . . . there exists a unique nonselected dyadic cube Q(k)x of generation

k that contains x. Then for each k ≥ 0, we have

∥∥∥∥∥1

|Q(k)x |

∫

Q(k)x

F(y)dy

∥∥∥∥∥B1

≤ 1

|Q(k)x |

∫

Q(k)x

‖F(y)‖B1dy≤ γα.

The intersection of the closures of the cubes Q(k)x is the singleton x. Using Corol-

lary 2.1.16, we deduce that for almost all x ∈ Rn \⋃ j Q j we have

F(x) =m

∑i=1

χRi(x)ui =

m

∑i=1

limk→∞

(1

|Q(k)x |

∫

Q(k)x

χRi(y)dy

)ui = lim

k→∞

1

|Q(k)x |

∫

Q(k)x

F(y)dy .

Since these averages are at most γα , we conclude that ‖F‖B1≤ γα almost every-

where on Rn \⋃ j Q j, hence ‖G‖B1≤ γα a.e. on this set. This proves (5.6.9).

By assumption we have

‖T (G)‖L∞(Rn,B) ≤ B⋆‖G‖L∞(Rn,B) ≤ 2nγαB⋆ = α/2 .

Then the set

x ∈ Rn : ‖T (G)(x)‖B2> α/2

is null and we have

∣∣x ∈ Rn : ‖T (F)(x)‖B2> α

∣∣≤∣∣x ∈ Rn : ‖T (B)(x)‖B2

> α/2∣∣.

Let Q∗j = 2√

nQ j. We have

∣∣x ∈Rn : ‖T (B)(x)‖B2> α/2

∣∣

≤∣∣⋃

j

Q∗j∣∣+

∣∣x /∈⋃

j

Q∗j : ‖T (B)(x)‖B2> α/2

∣∣

≤ (2√

n)n

γ

‖F‖L1(Rn,B1)

α+

2

α

∫

(∪ jQ∗j )

c‖T (B)(x)‖B2

dx

≤ (2√

n)n

γ

‖F‖L1(Rn,B1)

α+

2

α ∑j

∫

(Q∗j )c‖T (B j)(x)‖B2

dx,

since B = ∑ j B j. It suffices to estimate the last sum. Denoting by y j is the center of

the cube Q j and using the fact that B j has mean value zero over Q j, we write

∑j

∫

(Q∗j )c

∥∥T (B j)(x)∥∥

B2dx

= ∑j

∫

(Q∗j )c

∥∥∥∥∫

Q j

(K(x− y)−K(x− y j)

)(B j(y))dy

∥∥∥∥B2

dx

≤ ∑j

∫

Q j

∥∥B j(y)∥∥

B1

∫

(Q∗j )c

∥∥K(x− y)−K(x− y j)∥∥

B1→B2dxdy

≤ ∑j

∫

Q j

∥∥B j(y)∥∥

B1

∫

|x−y j |≥2|y−y j|

∥∥K(x− y)−K(x− y j)∥∥

B1→B2dxdy

≤ A∑j

‖B j‖L1(Q j ,B1)

≤ 2n+1A‖F‖L1(Rn,B1),

where we used the fact that |x−y j| ≥ 2|y−y j| for all x /∈Q∗j and y ∈Q j and (5.6.2).

Consequently,

∣∣x ∈ Rn : ‖T ( f )(x)‖B2> α

∣∣ ≤ (2√

n)n

γ

‖F‖L1(Rn,B1)

α+

2

α2n+1A‖F‖L1(Rn,B1)

=((2√

n)n2n+1B⋆+2n+1A)‖F‖L1(Rn,B1)

α

≤ C′n (A+B⋆)‖F‖L1(Rn,B1)

α,

where C′n = (2√

n)n2n+1 +2n+1. Thus T has an extension that maps L1(Rn,B1) to

L1,∞(Rn,B2) with constant Cn(A+B⋆). By interpolation (Exercise 5.5.3 (b)) it has

an extension that satisfies (5.6.7).

Case 2: 1 < r < ∞. We fix F = ∑mi=1 χRi

ui in Q⊗B1 and we notice that for each

x ∈Rn we have ‖F(x)‖B1=∑m

i=1 χRi(x)‖ui‖B1

. Thus the function x → ‖F(x)‖B1is

a finite linear combination of characteristic functions of disjoint dyadic cubes. We

prove the weak type estimate (5.6.6) by applying the Calderon–Zygmund decompo-

sition to the function x → ‖F(x)‖B1defined on Rn. Then we decompose F = G+B,

where G and B satisfy properties analogous to the case r =∞. The new ingredient in

this case is that the set

x ∈ Rn : ‖T (G)(x)‖B2> α/2

is not null but its measure

can be estimated as follows:

∣∣x ∈ Rn : ‖T (G)(x)‖B2> α/2

∣∣≤(

2B⋆

α

)r

‖G‖rL2(Rn,B1)

≤ 2

α‖F‖L1(Rn,B1)

,

where the first inequality is a consequence of the boundedness of T on Lr and the

second is obtained by combining (5.6.8) and (5.6.9). Combining this estimate for

the good function with the one for the bad function obtained in the preceding case,

it follows that T has an extension that satisfies (5.6.6), i.e., it maps T : L1(Rn,B1)to L1,∞(Rn,B2) with constant C′n (A+B⋆), where C′n = 2+(2

√n)n2n+1 +2n+1.

Next we interpolate between the estimates T : L1(Rn,B1) → L1,∞(Rn,B2)and T : Lr(Rn,B1) → Lr(Rn,B2). Using Exercise 5.5.3 (b) and the fact that

(p−1)−1/p ≤ (p−1)−1 when 1 r via duality. Since K(x) is an operator from B1 to B2,

its adjoint K∗(x) is an operator from B∗2 to B∗

1 . Obviously K∗(x) and K(x) have the

same norm, so (5.6.1) holds. For the same reason, condition (5.6.2) also holds forK∗, Finally condition (5.6.3) also holds since for any εk ↓ as k→ ∞ we have

∥∥∥∥∫

εk≤|y|≤1

K∗(y)dy− K0∗∥∥∥∥

B∗2→B∗

1

=

∥∥∥∥∫

εk≤|y|≤1

K(y)dy− K0

∥∥∥∥B1→B2

→ 0 .

Let T ′ be the Banach-valued operator with kernel K∗(−x). Clearly T ′ is well

defined on C ∞0 ⊗B∗

2 . For F(y) =∑mi=1 fi(y)w

∗i in C ∞

0 ⊗B∗2 and G(z) =∑l

j=1 g j(z)v j

in C ∞0 ⊗B1 we prove the following duality relation

∫

Rn

⟨T ′(F)(x),G(x)

⟩dx =

∫

Rn

⟨F(z),T (G)(z)

⟩dz . (5.6.11)

Indeed, for each index i ∈ 1, . . . ,m and j ∈ 1, . . . , l we have

∫

Rn

⟨limk→∞

∫

|y|≥εk

K∗(−y)( fi(x− y)w∗i )dy,g j(x)v j

⟩dx

= limk→∞

∫

|y|≥εk

∫

Rn

⟨K∗(−y)( fi(x− y)w∗i ),g j(x)v j

⟩dxdy

= limk→∞

∫

|y|≥εk

∫

Rn

⟨K∗(−y)( fi(z)w

∗i ),g j(z+ y)v j

⟩dzdy

= limk→∞

∫

|y|≥εk

∫

Rn

⟨fi(z)w

∗i , K(−y)(g j(z+ y)v j)

⟩dzdy

=∫

Rn

⟨fi(z)w

∗i , lim

k→∞

∫

|y|≥εk

K(y)(g j(z− y)v j)dy⟩

dz ,

proving (5.6.11), provided we can justify the interchange the x (or z)-integral with the

y-integral paired with the limit. These justifications can be given using the definition

in (5.6.4). For the part of the y-integral where |y| ≥ 1 the interchange is easily justi-

fied in view of the absolute convergence of the double integral and (5.5.21). For the

part of the y-integral where |y| ≤ 1 we introduce the operators K0∗ and K0 and we use

the facts that |g j(z− y)− g j(z)| ≤ ‖∇g j‖L∞ |y| and | fi(x− y)− fi(x)| ≤ ‖∇ fi‖L∞ |y|together with the assumption ‖K(y)‖B1→B2

≤ A|y|−n and (5.5.21) to obtain the ab-

solute convergence of the double integral, and thus justify the interchange.

We claim that T ′ is bounded from Lr′(Rn,B∗2) to Lr′(Rn,B∗

1). Indeed, to verify

this assertion, we fix F in C ∞0 ⊗B∗

2 and use Proposition 5.5.7 (b). Using (5.6.11),

for each G ∈ C ∞0 ⊗B1, we write

∣∣∣∫

Rn

⟨T ′(F)(x),G(x)

⟩dx

∣∣∣ =∣∣∣∫

Rn

⟨F(x),T (G)(x)

⟩dx

∣∣∣

≤∫

Rn

∥∥F(x)∥∥

B∗2

∥∥T (G)(x)∥∥

B2dx

≤ ‖F‖Lr′ (Rn,B∗

2)

∥∥T (G)∥∥

Lr(Rn,B2)

≤ ‖F‖Lr′ (Rn,B∗

2)B⋆ ‖G‖Lr(Rn,B1) ,

so taking the supremum over all G ∈ C ∞0 ⊗B1 with ‖G‖Lr(Rn,B1) ≤ 1 we deduce

that ∥∥T ′(F)∥∥

Lr′ (Rn,B∗1)≤ B⋆ ‖F‖Lr′ (Rn,B∗

2).

Collecting these facts, we have that T ′ is associated with a kernel K∗(−x) which

satisfies (5.6.1), (5.6.2), and (5.6.3) (with K0∗ in place of K0), and moreover it has a

bounded extension that maps Lr′(Rn,B∗2) to Lr′(Rn,B∗

1). The Calderon-Zygmund

decomposition in the vector-valued setting (discussed in the first paragraph of the

proof) yields that T ′ has an extension that satisfies

∥∥T ′(F)∥∥

L1,∞(Rn,B∗1)≤C′n(A+B⋆)‖F‖L1(Rn,B∗

2).

Using interpolation (Exercise 5.5.3 (b)) and the fact that (p′−1)1/p′ ≤ p, we obtain

that for 1 < p′ < r′, T ′ has an extension on Lp′(Rn,B∗2) that satisfies

∥∥T ′(F)∥∥

Lp′ (Rn,B∗1)≤Cn p(A+B⋆)‖F‖Lp′ (Rn,B∗

2). (5.6.12)

Let F =∑mi=1ϕi ui be in the dense subspace C ∞

0 ⊗B1 of Lp(Rn,B1). We observe

that ‖T (F)‖Lp(Rn,B2) < ∞. Indeed, all ϕi are supported in |x| ≤ R, then for |x| ≥ 2R

we have

∥∥∥∥∫

|y|≤R

K(x− y)( m

∑i=1

ϕi ui

)dy

∥∥∥∥B2

≤ A( |x|

2

)−n m

∑i=1

‖ϕi‖L1‖ui‖B1(5.6.13)

which is integrable to the power p > 1 in the region |x| ≥ 2R. Also using the defi-

nition in (5.6.4) we see that the expression on the left in (5.6.13) is bounded, hence

integrable to the power p in the region |x| ≤ 2R. For a fixed r < p < ∞, we are now

able to apply Proposition 5.5.7 (a) to write

∥∥T (F)∥∥

Lp(Rn,B2)≤ sup‖G‖

Lp′ (Rn,B∗2)≤1

∣∣∣∣∫

Rn

⟨G(x),T (F)(x)

⟩dx

∣∣∣∣

= sup‖G‖

Lp′ (Rn,B∗2)≤1

∣∣∣∣∫

Rn

⟨T ′(G)(x),F(x)

⟩dx

∣∣∣∣

≤ sup‖G‖

Lp′ (Rn,B∗2)≤1

∥∥T ′(G)∥∥

Lp′ (Rn,B∗1)‖F‖Lp(Rn,B1)

≤Cn p(A+B⋆)‖F‖Lp(Rn,B1) ,

where we used (5.6.12). This combined with (5.6.10) implies the required conclu-

sion whenever r < ∞ and p ∈(1,min(r,2)

)∪ (r,∞). The remaining p’s follow by

interpolation (Exercise 5.5.3 (a)).

5.6.2 Applications

We proceed with some applications. An important consequence of Theorem 5.6.1 is

the following:

Corollary 5.6.2. Let A,B > 0 and let Wj be a sequence of tempered distributions

on Rn whose Fourier transforms are uniformly bounded functions (i.e., |Wj| ≤ B).

Suppose that for each j, Wj coincides with a function K j on Rn \0 that satisfies

|K j(x)| ≤ A |x|−n , x = 0, (5.6.14)

limεk→0

∫

|x|≥εk

K j(x)dx = L j , (5.6.15)

for some complex constant L j, and

supy∈Rn\0

∫

|x|≥2|y|sup

j

|K j(x− y)−K j(x)|dx≤ A . (5.6.16)

Then there are constants Cn,C′n > 0 such that for all 1 < p,r < ∞ we have

∥∥∥(∑

j

|Wj ∗ f j|r)1

r∥∥∥

L1,∞≤C′n max(r,(r−1)−1)(A+B)

∥∥∥(∑

j

| f j|r)1

r∥∥∥

L1,

∥∥∥(∑

j

|Wj ∗ f j|r)1

r∥∥∥

Lp≤Cn c(p,r)(A+B)

∥∥∥(∑

j

| f j|r)1

r∥∥∥

Lp,

where c(p,r) = max(p,(p−1)−1)max(r,(r−1)−1).

Proof. Let Tj be the operator given by convolution with the distribution Wj. Clearly

Tj is L2 bounded with norm at most B. It follows from Theorem 5.3.3 that the Tj’s

are of weak type (1,1) and also bounded on Lr with bounds at most a dimensional

constant multiple of max(r,(r−1)−1)(A+B), uniformly in j. We set B1 =B2 = ℓr

and defineT ( f j j) = Wj ∗ f j j

for f j j ∈ Lr(Rn, ℓr). It is immediate to verify that T maps Lr(Rn, ℓr) to itself

with norm at most a dimensional constant multiple of max(r,(r−1)−1)(A+B). The

kernel of T is K in L(ℓr, ℓr) defined by

K(x)(t j j) = K j(x)t j j, t j j ∈ ℓr.

Obviously, we have

∥∥K(x− y)−K(x)∥∥ℓr→ℓr ≤ sup

j

|K j(x− y)−K j(x)| ,

and therefore condition (5.6.3) holds for K as a consequence of (5.6.16). Moreover,

(5.6.1) and (5.6.2) with K0 = L j j are also valid for this K, in view of assumptions

(5.6.14) and (5.6.15). The desired conclusion follows from Theorem 5.6.1.

If all the Wj’s are equal, we obtain the following corollary, which contains in

particular the inequality (5.5.16) mentioned earlier.

Corollary 5.6.3. Let W be an element of S ′(Rn) whose Fourier transform is a func-

tion bounded in absolute value by some B > 0. Suppose that W coincides with some

locally integrable function K on Rn \0 that satisfies

|K(x)| ≤ A |x|−n , x = 0,

limεk→0

∫

εk≤|x|≤1K(x)dx = L ,

and

supy∈Rn\0

∫

|x|≥2|y||K(x− y)−K(x)|dx≤ A . (5.6.17)

Let T be the operator given by convolution with W. Then there exist constants

Cn,C′n > 0 such that for all 1 < p,r < ∞ we have that

∥∥∥(∑

j

|T ( f j)|r)1

r∥∥∥

L1,∞≤C′n max(r,(r−1)−1)(A+B)

∥∥∥(∑

j

| f j|r)1

r∥∥∥

L1,

∥∥∥(∑

j

|T ( f j)|r)1

r∥∥∥

Lp≤Cnc(p,r)(A+B)

∥∥∥(∑

j

| f j|r)1

r∥∥∥

Lp,

where c(p,r)=max(p,(p−1)−1)max(r,(r−1)−1). In particular, these inequalities

are valid for the Hilbert transform and the Riesz transforms.

Interestingly enough, we can use the very statement of Theorem 5.6.1 to obtain

its corresponding vector-valued version.

Proposition 5.6.4. Let let 1 < p,r < ∞ and let B1 and B2 be two Banach spaces.

Suppose that T given by (5.6.4) is a bounded linear operator from Lr(Rn,B1) to

Lr(Rn,B2) with norm B = B(r). Also assume that for all x ∈ Rn \ 0, K(x) is a

bounded linear operator from B1 to B2 that satisfies conditions (5.6.1) , (5.6.2),

(5.6.3) for some A > 0 and K0 ∈ L(B1,B2). Then there exist positive constants

Cn,C′n such that for all B1-valued functions Fj we have

∥∥∥(∑

j

∥∥T (Fj)∥∥r

B2

)1r∥∥∥

L1,∞(Rn)≤C′n(A+B)

∥∥∥(∑

j

∥∥Fj

∥∥r

B1

)1r∥∥∥

L1(Rn),

∥∥∥(∑

j

∥∥T (Fj)∥∥r

B2

)1r∥∥∥

Lp(Rn)≤Cn(A+B)c(p)

∥∥∥(∑

j

∥∥Fj

∥∥r

B1

)1r∥∥∥

Lp(Rn),

where c(p) = max(p,(p−1)−1).

Proof. Let us denote by ℓr(B1) the Banach space of all B1-valued sequences u j j

that satisfy ∥∥u j j

∥∥ℓr(B1)

=(∑

j

∥∥u j

∥∥r

B1

)1r < ∞.

Now consider the operator S defined on Lr(Rn, ℓr(B1)) by

S(Fj j) = T (Fj) j .

It is obvious that S maps Lr(Rn, ℓr(B1)) to Lr(Rn, ℓr(B2)) with norm at most B.

Moreover, S has kernel K(x) ∈ L(ℓr(B1), ℓr(B2)) given by

K(x)(u j j) = K(x)(u j) j,

where K is the kernel of T . It is not hard to verify that for x ∈ Rn \0 we have

∥∥K(x)∥∥ℓr(B1)→ℓr(B2)

=∥∥K(x)

∥∥B1→B2

,

hence for x = y ∈ Rn we also have

∥∥K(x− y)− K(x)∥∥ℓr(B1)→ℓr(B2)

=∥∥K(x− y)−K(x)

∥∥B1→B2

.

Moreover, if we define K0 ∈ L(ℓr(B1), ℓ

r(B2))

by

K0(u j j) =K0(y)(u j)

j.

for u j j ∈ ℓr(B1), then we have

limk→∞

∫

εk≤|y|≤1K(y)dy = K0,

in L(ℓr(B1), ℓ

r(B2)).

We conclude that K satisfies conditions (5.6.1) , (5.6.2), (5.6.3). Hence the oper-

ator S associate with K satisfies the conclusion of Theorem 5.6.1, that is, the desired

inequalities for T .

5.6.3 Vector-Valued Estimates for Maximal Functions

Next, we discuss applications of vector-valued inequalities to some nonlinear opera-

tors. We fix an integrable functionΦ on Rn and for t > 0 defineΦt(x) = t−nΦ(t−1x).We suppose that Φ satisfies the following regularity condition:

supy∈Rn\0

∫

|x|≥2|y|supt>0

|Φt(x− y)−Φt(x)|dx = AΦ < ∞ . (5.6.18)

We consider the maximal operator

MΦ( f )(x) = supt>0

|( f ∗Φt)(x)|

defined for f in L1 +L∞. We are interested in obtaining Lp estimates for MΦ . We

observe that the trivial estimate

∥∥MΦ( f )∥∥

L∞≤ ‖Φ‖L1‖ f‖L∞ (5.6.19)

holds when p = ∞. It is natural to set

B1 = C and B2 = L∞(R+)

and view MΦ as the linear operator f → f ∗Φδδ>0 that maps B1-valued functions

to B2-valued functions.

To do this precisely, for each x ∈ Rn we define a bounded linear operator KΦ(x)from B1 to B2 by setting for c ∈ C

KΦ(x)(c) = c Φδ (x)δ∈R+ .

Clearly we have ∥∥KΦ(x)∥∥

C→L∞(R+)= sup

δ>0

|Φδ (x)| .

Now (5.6.18) implies condition (5.6.2) for the kernel KΦ . Also condition (5.6.1)

holds (for some A < depending on n) since

supδ>0

|Φδ (x)| ≤ A |x|−n

and also condition (5.6.3) holds since for every δ > 0 we have

limε→0

∫

ε≤|y|≤1Φδ (y)dy =

∫

|y|≤1Φδ (y)dy .

We also define a B2-valued linear operator acting on complex-valued functions

on Rn byMΦ( f ) = f ∗KΦ = f ∗Φδδ∈R+ .

Obvisouly MΦ maps L∞(Rn,B1) to L∞(Rn,B2) with norm at most ‖Φ‖L1 .

Applying Theorem 5.6.1 with r = ∞ we obtain for 1 < p < ∞,

∥∥MΦ( f )∥∥

Lp(Rn,B2)≤Cn max(p,(p−1)−1)

(AΦ +‖Φ‖L1

)∥∥ f∥∥

Lp(Rn), (5.6.20)

which can be immediately improved to

∥∥MΦ( f )∥∥

Lr(Rn,B2)≤Cn max(1,(r−1)−1)

(AΦ +‖Φ‖L1

)∥∥ f∥∥

Lr(Rn)(5.6.21)

via interpolation with estimate (5.6.19) for all 1 < r < ∞.

Next we use estimate (5.6.21) to obtain vector-valued estimates for the sublinear

operator MΦ .

Corollary 5.6.5. Let Φ be an integrable function on Rn that satisfies (5.6.18). Then

there exist dimensional constants Cn and C′n such that for all 1 < p,r < ∞ the fol-

lowing vector-valued inequalities are valid:

∥∥∥(∑

j

|MΦ( f j)|r)1

r∥∥∥

L1,∞≤C′nc(r)

(AΦ+‖Φ‖L1

)∥∥∥(∑

j

| f j|r)1

r∥∥∥

L1, (5.6.22)

where c(r) = 1+(r−1)−1, and

∥∥∥(∑

j

|MΦ( f j)|r)1

r∥∥∥

Lp≤Cnc(p,r)

(AΦ+‖Φ‖L1

)∥∥∥(∑

j

| f j|r)1

r∥∥∥

Lp, (5.6.23)

where c(p,r) =(1+(r−1)−1

)(p+(p−1)−1

).

Proof. We set B1 = C and B2 = L∞(R+). We use estimate (5.6.21) as a start-

ing point in Proposition 5.6.4, which immediately yields the required conclusions

(5.6.22) and (5.6.23).

Similar estimates hold for the Hardy–Littlewood maximal operator.

Theorem 5.6.6. For 1 < p,r < ∞ the Hardy–Littlewood maximal function M satis-

fies the vector-valued inequalities

∥∥∥(∑

j

|M( f j)|r)1

r∥∥∥

L1,∞≤C′n

(1+(r−1)−1

)∥∥∥(∑

j

| f j|r)1

r∥∥∥

L1, (5.6.24)

∥∥∥(∑

j

|M( f j)|r)1

r∥∥∥

Lp≤Cn c(p,r)

∥∥∥(∑

j

| f j|r)1

r∥∥∥

Lp, (5.6.25)

where c(p,r) =(1+(r−1)−1

)(p+(p−1)−1

).

Proof. Let us fix a positive radial symmetrically decreasing Schwartz functionΦ on

Rn that satisfies Φ(x)≥ 1 when |x| ≤ 1. Then the Hardy–Littlewood maximal func-

tion M( f ) is pointwise controlled by a constant multiple of the function MΦ(| f |).

In view of Corollary 5.6.5, it suffices to check that for such a Φ , (5.6.18) holds. First

observe that in view of the decreasing character of Φ , we have

supj

| f | ∗Φ2 j ≤MΦ(| f |)≤ 2n supj

| f | ∗Φ2 j ,

and for this reason we choose to work with the easier dyadic maximal operator

MdΦ( f ) = sup

j

| f ∗Φ2 j | .

We observe the validity of the simple inequalties

2−n M( f )≤M( f )≤MΦ(| f |)≤ 2nMdΦ(| f |) . (5.6.26)

If we can show that

supy∈Rn\0

∫

|x|≥2|y|supj∈Z

|Φ2 j(x− y)−Φ2 j(x)|dx =Cn < ∞ , (5.6.27)

then (5.6.22) and (5.6.23) are satisfied with MdΦ replacing MΦ . We therefore turn

our attention to (5.6.27). We have

∫

|x|≥2|y|supj∈Z

|Φ2 j(x− y)−Φ2 j(x)|dx

≤ ∑j∈Z

∫

|x|≥2|y||Φ2 j(x− y)−Φ2 j(x)|dx

≤ ∑2 j>|y|

∫

|x|≥2|y|

|y| |∇Φ(

x−θy

2 j

)|

2(n+1) jdx+ ∑

2 j≤|y|

∫

|x|≥2|y|(|Φ2 j(x− y)|+ |Φ2 j(x)|)dx

≤ ∑2 j>|y|

∫

|x|≥2|y|

|y|2(n+1) j

CN dx

(1+ |2− j(x−θy)|)N+2 ∑

2 j≤|y|

∫

|x|≥|y||Φ2 j(x)|dx

≤ ∑2 j>|y|

∫

|x|≥2|y|

|y|2(n+1) j

CN

(1+ |2− j−1x|)Ndx+2 ∑

2 j≤|y|

∫

|x|≥2− j |y||Φ(x)|dx

≤ ∑2 j>|y|

∫

|x|≥2− j|y|

|y|2 j

CN

(1+ |x|)Ndx+2 ∑

2 j≤|y|CN(2

− j|y|)−N

≤CN ∑2 j>|y|

|y|2 j

+CN

≤ 3CN ,

where CN > 0 depends on N > n, θ ∈ [0,1], and |x−θy| ≥ |x|/2 when |x| ≥ 2|y|.Now apply (5.6.22) and (5.6.23) to Md

Φ and use (5.6.26) to obtain the desired

vector-valued inequalities.

Remark 5.6.7. Observe that (5.6.24) and (5.6.25) also hold for r = ∞. These end-

point estimates can be proved directly by observing that

supj

M( f j)≤M(supj

| f j|) .

The same is true for estimates (5.6.22) and (5.6.23). Finally, estimates (5.6.25) and

(5.6.23) also hold for p = ∞.

Exercises

5.6.1. (a) For all j ∈ Z, let I j be an interval in R and let Tj be the operator given

on the Fourier transform by multiplication by the characteristic function of I j. Prove

that there exists a constant C > 0 such that for all 1 < p,r < ∞ and for all square

integrable functions f j on R we have

∥∥∥(∑

j

|Tj( f j)|r)1

r∥∥∥

Lp(R)≤C max

(r,

1

r−1

)max

(p,

1

p−1

)∥∥∥(∑

j

| f j|r)1

r∥∥∥

Lp(R),

∥∥∥(∑

j

|Tj( f j)|r)1

r∥∥∥

L1,∞(R)≤C max

(r,

1

r−1

)∥∥∥(∑

j

| f j|r)1

r∥∥∥

L1(R).

(b) Let R j be arbitrary rectangles on Rn with sides parallel to the axes and let S j be

the operators given on the Fourier transform by multiplication by the characteristic

functions of R j. Prove that there exists a dimensional constant Cn < ∞ such that for

all indices 1 < p,r <∞ and for all square integrable functions f j in Lp(Rn) we have

∥∥∥(∑

j

|S j( f j)|r)1

r∥∥∥

Lp(Rn)≤Cn max

(r,

1

r−1

)n

max(

p,1

p−1

)n∥∥∥(∑

j

| f j|r)1

r∥∥∥

Lp(Rn).

[Hint: Part (a): Use Theorem 5.5.1 and the identity Tj =

i2

(MaHM−a−MbHM−b

),

if I j is χ(a,b), where Ma( f )(x) = f (x)e2πiax and H is the Hilbert transform. Part (b):

Apply the result in part (a) in each variable.]

5.6.2. Let (T,dµ) be a σ -finite measure space. For every t ∈ T , let R(t) be a rect-

angle in Rn with sides parallel to the axes such that the map t → R(t) is measurable.

Then there is a constant Cn > 0 such that for all 1 < p < ∞ and for all families of

square integrable functions ftt∈T on Rn such that t → ft(x) is measurable for all

x ∈ Rn we have

∥∥∥∥(∫

T|( ftχR(t))

∨|2 dµ(t)

)12∥∥∥∥

Lp

≤Cn max(p,(p−1)−1)n

∥∥∥∥(∫

T| ft |2 dµ(t)

)12∥∥∥∥

Lp

,

[Hint: When n = 1 reduce matters to an Lp(L2(T,dµ),L2(T,dµ)) inequality for the

Hilbert transform, via the hint in the preceding exercise. Verify the inequality p = 2

and then use Theorem 5.6.1 for the other p’s. Obtain the n-dimensional inequality

by iterating the one-dimensional.]

5.6.3. Let Φ be a function on Rn that satisfies supx∈Rn |x|n|Φ(x)| ≤ A and

∫

Rn|Φ(x− y)−Φ(x)|dx≤ η(y),

∫

|x|≥R|Φ(x)|dx≤ η(R−1) ,

for all R ≥ 1, where η is a continuous increasing function on [0,2] that satisfies

η(0) = 0 and∫ 2

0η(t)

tdt < ∞ .

(a) Prove that (5.6.27) holds.

(b) Show that if Φ lies in L1(Rn), then the maximal function f → sup j∈Z | f ∗Φ2 j |maps Lp(Rn) to itself for 1 < p≤ ∞.[Hint: Part (a): Modify the calculation in the proof of Theorem 5.6.6. Part (b): Use

Theorem 5.6.1 with r = ∞.]

5.6.4. (a) On R, take f j = χ[2 j−1,2 j ] to prove that inequality (5.6.25) fails when p=∞and 1 < r < ∞.

(b) Again on R, take N > 2 and f j = χ[ j−1

N , jN ]

for j = 1,2, . . . ,N to prove that (5.6.25)

fails when 1 < p < ∞ and r = 1.

5.6.5. Let K be an integrable function on the real line and assume that the operator

f → f ∗K is bounded on Lp(R) for some 1 < p < ∞. Prove that the vector-valued

inequality ∥∥∥(∑

j

|K ∗ f j|q)1

q∥∥∥

Lp≤Cp,q

∥∥∥(∑

j

| f j|q)1

q∥∥∥

Lp

may fail in general when q < 1.[Hint: Take K = χ[−1,1] and f j = χ

[ j−1N , j

N ]for 1≤ j ≤ N.

]

5.6.6. Let Q j j be a countable collection of cubes in Rn with disjoint interiors.

Let c j be the center of the cube Q j and d j its diameter. For ε > 0, define the

Marcinkiewicz function associated with the family Q j j as follows:

Mε(x) =∑j

dn+εj

|x− c j|n+ε +dn+εj

.

Prove that for some constants Cn,ε ,p and Cn,ε one has

∥∥Mε

∥∥Lp ≤Cn,ε ,p

(∑

j

|Q j|) 1

p, p >

n

n+ ε,

∥∥Mε

∥∥L

nn+ε ,∞

≤Cn,ε

(∑

j

|Q j|) n+ε

n,

and consequently∫

Rn Mε(x)dx≤Cn,ε ∑ j |Q j|.

[Hint: Verify that

dn+εj

|x− c j|n+ε +dn+εj

≤CM(χQ j)(x)

n+εn

and use Theorem 5.6.6.]

HISTORICAL NOTES

The Lp boundedness of the conjugate function on the circle was announced in 1924 by Riesz

[292], but its first proof appeared three years later in [294]. In view of the identification of the

Hilbert transform with the conjugate function, the Lp boundedness of the Hilbert transform is also

attributed to M. Riesz. Riesz’s proof was first given for p = 2k, k ∈ Z+, via an argument similar to

that in the proof of Theorem 4.1.7. For p = 2k this proof relied on interpolation and was completed

with the simultaneous publication of Riesz’s article on interpolation of bilinear forms [293]. The

weak type (1,1) property of the Hilbert transform is due to Kolmogorov [197]. Additional proofs

of the boundedness of the Hilbert transform have been obtained by Stein [350], Loomis [230], and

Calderon [41]. The proof of Theorem 5.1.7, based on identity (5.1.23), is a refinement of a proof

given by Cotlar [75].

The norm of the conjugate function on Lp(T1), and consequently that of the Hilbert transform

on Lp(R), was shown by Gohberg and Krupnik [129] to be cot(π/2p) when p is a power of 2. Du-

ality gives that this norm is tan(π/2p) for 1 < p≤ 2 whenever p′ is a power of 2. Pichorides [282]

extended this result to all 1 < p < ∞ by refining Calderon’s proof of Riesz’s theorem. This result

was also independently obtained by B. Cole (unpublished). The direct and simplified proof for the

Hilbert transform given in Exercise 5.1.12 is in Grafakos [130]. The norm of the operators 12(I± iH)

for real-valued functions was found to be 12

[min(cos(π/2p),sin(π/2p))

]−1by Verbitsky [366] and

later independently by Essen [108]. The norm of the same operators for complex-valued functions

was shown to be equal to [sin(π/p)]−1 by Hollenbeck and Verbitsky [156]. Exact formulas for the

Lp norm, 1 ≤ p < ∞ of the Hilbert transform acting on a characteristic function were obtained by

Laeng [211]. The best constant in the weak type (1,1) estimate for the Hilbert transform is equal

to (1+ 132 + 1

52 + · · ·)(1− 132 + 1

52 − ·· ·)−1 as shown by Davis [91] using Brownian motion; an

alternative proof was later obtained by Baernstein [17]. Iwaniec and Martin [175] showed that the

norms of the Riesz transforms on Lp(Rn) coincide with that of the Hilbert transform on Lp(R) for

1 < p < ∞.

Operators of the kind TΩ as well as the stopping-time decomposition of Theorem 5.3.1 were

introduced by Calderon and Zygmund [46]. In the same article, Calderon and Zygmund used this

decomposition to prove Theorem 5.3.3 for operators of the form TΩ when Ω satisfies a certain

weak smoothness condition. The more general condition (5.3.12) first appeared in Hormander’s

article [159]. A more flexible condition sufficient to yield weak type (1,1) bounds is contained

in the article of Duong and McIntosh [104]. Theorems 5.2.10 and 5.2.11 are also due to Calderon

and Zygmund [48]. The latter article contains the method of rotations. Algebras of operators of the

form TΩ were studied in [49]. For more information on algebras of singular integrals see the article

of Calderon [44]. Theorem 5.4.1 is due to Benedek, Calderon, and Panzone [22], while Example

5.4.2 is taken from Muckenhoupt [259]. Theorem 5.4.5 is due to Riviere [296]. A weaker version

of this theorem, applicable for smoother singular integrals such as the maximal Hilbert transform,

was obtained by Cotlar [75] (Theorem 5.3.4). Improvements of the main inequality in Theorem

5.3.4 for homogeneous singular integrals were obtained by Mateu and Verdera [245] and Mateu,

Orobitg, and Verdera [244]. For a general overview of singular integrals and their applications, one

may consult the expository article of Calderon [43].

Part (a) of Theorem 5.5.1 is due to Marcinkiewicz and Zygmund [242], although the case p = q

was proved earlier by Paley [273] with a larger constant. The values of r for which a general linear

operator of weak or strong type (p,q) admits bounded ℓr extensions are described in Rubio de

Francia and Torrea [304]. The Lp and weak Lp spaces in Theorem 5.5.1 can be replaced by general

Banach lattices, as shown by Krivine [206] using Grothendieck’s inequality. Hilbert-space-valued

estimates for singular integrals were obtained by Benedek, Calderon, and Panzone [22]. Other

operator-valued singular integral operators were studied by Rubio de Francia, Ruiz, and Torrea

[303]. Banach-valued singular integrals are studied in great detail in the book of Garcıa-Cuerva

and Rubio de Francia [122], which provides an excellent presentation of the subject. The ℓr-valued

estimates (5.5.16) for the Hilbert transform were first obtained by Boas and Bochner [28]. The

corresponding vector-valued estimates for the Hardy–Littlewood maximal function in Theorem

5.6.6 are due to Fefferman and Stein [115]. Conditions of the form (5.6.18) have been applied to

several situations and can be traced in Zo [386].

The sharpness of the logarithmic condition (5.2.24) was indicated by Weiss and Zygmund [372],

who constructed an example of an integrable function Ω with vanishing integral on S1 satisfying∫

Sn−1 |Ω(θ)| log+ |Ω(θ)|(

log(2+ log(2+ |Ω(θ)|)))−δ

dθ = ∞ for all δ > 0 and of a continuous

function in Lp(R2) for all 1 < p < ∞ such that limsupε→0 |T(ε)Ω ( f )(x)|= ∞ for almost all x ∈ R2.

The proofs of Theorems 5.2.10 and 5.2.11 can be modified to give that if Ω is in the Hardy space

H1 of Sn−1, then TΩ and T(∗)Ω map Lp to Lp for 1 < p < ∞. For TΩ this fact was proved by Connett

[72] and independently by Ricci and Weiss [289]; for T(∗)Ω this was proved by Fan and Pan [110]

and independently by Grafakos and Stefanov [139]. The latter authors [138] also obtained that the

logarithmic condition ess.sup|ξ |=1

∫Sn−1 |Ω(θ)|(log 1

|ξ ·θ | |)1+α dθ <∞, α > 0, implies Lp bounded-

ness for TΩ and T(∗)Ω for some p = 2. See also Fan, Guo, and Pan [109] as well as Ryabogin and

Rubin [308] for extensions. Examples of functions Ω for which TΩ maps Lp to Lp for a certain

range of p’s but not for other ranges of p’s is given in Grafakos, Honzık, and Ryabogin [132]. A

different example of this sort was provided later by Honzık [158]; the range of p’s for which bound-

edness holds are different for these examples. Honzık [157] also constructed a delicate example of

an integrable function Ω with mean value zero over S1 such that TΩ is bounded on L2(R) but T(∗)Ω

is not.

The relatively weak condition |Ω | log+ |Ω | ∈ L1(Sn−1) also implies weak type (1,1) bound-

edness for operators TΩ . This was obtained by Seeger [317] and later extended by Tao [355] to

situations in which there is no Fourier transform structure. Earlier partial results are in Christ and

Rubio de Francia [63] and in the simultaneous work of Hofmann [155], both inspired by the work

of Christ [60]. Soria and Sjogren [324] showed that for arbitrary Ω in L1(Sn−1), TΩ is weak type

(1,1) when restricted to radial functions. Examples due to Christ (published in [139]) indicate

that even for bounded functions Ω on Sn−1, TΩ may not map the endpoint Hardy space H1(Rn) to

L1(Rn). However, Seeger and Tao [318] have showed that TΩ always maps the Hardy space H1(Rn)to the Lorentz space L1,2(Rn) when |Ω |(log+ |Ω |)2 is integrable over Sn−1. This result is sharp in

the sense that for such Ω , TΩ may not map H1(Rn) to L1,q(Rn) when q < 2 in general. If TΩ maps

H1(Rn) to itself, Daly and Phillips [87] (in dimension n = 2) and Daly [86] (in dimensions n≥ 3)

showed that Ω must lie in the Hardy space H1(Sn−1). There are also results concerning the singu-

lar maximal operator MΩ ( f )(x) = supr>01

vnrn

∫|y|≤r | f (x− y)| |Ω(y)|dy, where Ω is an integrable

function on Sn−1 of not necessarily vanishing integral. Such operators were studied by Fefferman

[116], Christ [60], and Hudson [162]. An excellent treatment of several kinds of singular integral

operators with rough kernels is contained in the book of Lu, Ding, and Yan [234].

Chapter 6

Littlewood–Paley Theory and Multipliers

In this chapter we are concerned with orthogonality properties of the Fourier trans-

form. This orthogonality is easily understood on L2, but at this point it is not clear

how it manifests itself on other spaces. Square functions introduce a way to express

and quantify orthogonality of the Fourier transform on Lp and other function spaces.

The introduction of square functions in this setting was pioneered by Littlewood and

Paley, and the theory that subsequently developed is named after them. The extent

to which Littlewood–Paley theory characterizes function spaces is remarkable.

Historically, Littlewood–Paley theory first appeared in the context of one-dimen-

sional Fourier series and depended on complex function theory. With the develop-

ment of real-variable methods, the whole theory became independent of complex

methods and was extended to Rn. This is the approach that we follow in this chapter.

It turns out that the Littlewood–Paley theory is intimately related to the Calderon–

Zygmund theory introduced in the previous chapter. This connection is deep and

far-reaching, and its central feature is that one is able to derive the main results of

one theory from the other.

The thrust and power of the Littlewood–Paley theory become apparent in some of

the applications we discuss in this chapter. Such applications include the derivation

of certain multiplier theorems, that is, theorems that yield sufficient conditions for

bounded functions to be Lp multipliers. As a consequence of Littlewood–Paley the-

ory we also prove that the lacunary partial Fourier integrals∫|ξ |≤2N f (ξ )e2πix·ξ dξ

converge almost everywhere to an Lp function f on Rn.

6.1 Littlewood–Paley Theory

We begin by examining more closely what we mean by orthogonality of the Fourier

transform. If the functions f j defined on Rn have Fourier transforms f j supported in

disjoint sets, then they are orthogonal in the sense that



419

420 6 Littlewood–Paley Theory and Multipliers

∥∥∑j

f j

∥∥2

L2 =∑j

∥∥ f j

∥∥2

L2 . (6.1.1)

Unfortunately, when 2 is replaced by some p = 2 in (6.1.1), the previous quanti-

ties may not even be comparable, as we show in Examples 6.1.8 and 6.1.9. The

Littlewood–Paley theorem provides a substitute inequality to (6.1.1) expressing the

fact that certain orthogonality considerations are also valid in Lp(Rn).

6.1.1 The Littlewood–Paley Theorem

The orthogonality we are searching for is best seen in the context of one-dimensional

Fourier series (which was the setting in which Littlewood and Paley formulated

their result). The primary observation is that the exponential e2πi2kx oscillates half

as much as e2πi2k+1x and is therefore nearly constant in each period of the latter.

This observation was instrumental in the proof of Theorem 3.6.4, which implied in

particular that for all 1 < p < ∞ we have

∥∥∥N

∑k=1

ake2πi2kx∥∥∥

Lp[0,1]≈

( N

∑k=1

|ak|2) 1

2. (6.1.2)

In other words, we can calculate the Lp norm of ∑Nk=1 ake2πi2kx in almost a pre-

cise fashion to obtain (modulo multiplicative constants) the same answer as in the

L2 case. Similar calculations are valid for more general blocks of exponentials in

the dyadic range 2k + 1, . . . ,2k+1− 1, since the exponentials in each such block

behave independently from those in each previous block. In particular, the Lp inte-

grability of a function on T1 is not affected by the randomization of the sign of its

Fourier coefficients in the previous dyadic blocks. This is the intuition behind the

Littlewood–Paley theorem.

Motivated by this discussion, we introduce the Littlewood–Paley operators in the

continuous setting.

Definition 6.1.1. Let Ψ be an integrable function on Rn and j ∈ Z. We define the

Littlewood–Paley operator ∆ j associated withΨ by

∆ j( f ) = f ∗Ψ2− j ,

where Ψ2− j(x) = 2 jnΨ(2 jx) for all x in Rn. Thus we have Ψ2− j(ξ ) = Ψ(2− jξ ) for

all ξ in Rn. We note that whenever Ψ is a Schwartz function and f is a tempered

distribution, the quantity ∆ j( f ) is a well defined function.

These operators depend on the choice of the function Ψ ; in most applications

we chooseΨ to be a smooth function with compactly supported Fourier transform.

Observe that if Ψ is supported in some annulus 0 < c1 < |ξ | < c2 < ∞, then the

Fourier transform of ∆ j is supported in the annulus c12 j < |ξ | < c22 j; in other

6.1 Littlewood–Paley Theory 421

words, it is localized near the frequency |ξ | ≈ 2 j. Thus the purpose of ∆ j is to

isolate the part of frequency of a function concentrated near |ξ | ≈ 2 j.

The square function associated with the Littlewood–Paley operators ∆ j is

defined by

f →(∑j∈Z

|∆ j( f )|2) 1

2.

This quadratic expression captures the intrinsic orthogonality of the function f .

Theorem 6.1.2. (Littlewood–Paley theorem) Suppose that Ψ is an integrable C 1

function on Rn with mean value zero that satisfies

|Ψ(x)|+ |∇Ψ(x)| ≤ B(1+ |x|)−n−1 . (6.1.3)

Then there exists a constant Cn < ∞ such that for all 1 < p < ∞ and all f in Lp(Rn)we have

∥∥∥(∑j∈Z

|∆ j( f )|2) 1

2∥∥∥

Lp(Rn)≤CnBmax

(p,(p−1)−1

)∥∥ f∥∥

Lp(Rn). (6.1.4)

There also exists a C′n < ∞ such that for all f in L1(Rn) we have

∥∥∥(∑j∈Z

|∆ j( f )|2) 1

2∥∥∥

L1,∞(Rn)≤C′nB

∥∥ f∥∥

L1(Rn). (6.1.5)

Conversely, letΨ be a Schwartz function such that either Ψ(0) = 0 and

∑j∈Z

|Ψ(2− jξ )|2 = 1, for all ξ ∈ Rn \0, (6.1.6)

or Ψ is compactly supported away from the origin and

∑j∈Z

Ψ(2− jξ ) = 1, for all ξ ∈ Rn \0. (6.1.7)

Then there is a constant Cn,Ψ , such that for any f ∈S ′(Rn) with(∑ j∈Z |∆ j( f )|2

) 12

in Lp(Rn) for some 1 < p < ∞, there exists a unique polynomial Q such that the

tempered distribution f −Q coincides with an Lp function, and we have

∥∥ f −Q∥∥

Lp(Rn)≤Cn,Ψ Bmax

(p,(p−1)−1

)∥∥∥(∑j∈Z

|∆ j( f )|2) 1

2∥∥∥

Lp(Rn). (6.1.8)

Consequently, if g lies in Lp(Rn) for some 1 < p < ∞, then

∥∥g‖Lp(Rn) ≈∥∥∥(∑j∈Z

|∆ j(g)|2) 1

2∥∥∥

Lp(Rn).

Proof. We first prove (6.1.4) when p = 2. Using Plancherel’s theorem, we see that

(6.1.4) is a consequence of the inequality

∑j

|Ψ(2− jξ )|2 ≤CnB2 (6.1.9)

for some Cn <∞. Because of (6.1.3), Fourier inversion holds forΨ . Furthermore,Ψhas mean value zero and we may write

Ψ(ξ ) =∫

Rne−2πix·ξΨ(x)dx =

∫

Rn(e−2πix·ξ −1)Ψ(x)dx , (6.1.10)

from which we obtain the estimate

|Ψ(ξ )| ≤√

4π|ξ |∫

Rn|x| 1

2 |Ψ(x)|dx≤CnB|ξ | 12 . (6.1.11)

For ξ = (ξ1, . . . ,ξn) = 0, let j be such that |ξ j| ≥ |ξk| for all k ∈ 1, . . . ,n. Integrate

by parts with respect to ∂ j in (6.1.10) to obtain

Ψ(ξ ) =−∫

Rn(−2πiξ j)

−1e−2πix·ξ (∂ jΨ)(x)dx,

from which we deduce the estimate

|Ψ(ξ )| ≤√

n |ξ |−1∫

Rn|∇Ψ(x)|dx≤CnB|ξ |−1. (6.1.12)

We now break the sum in (6.1.9) into the parts where 2− j|ξ | ≤ 1 and 2− j|ξ | ≥ 1

and use (6.1.11) and (6.1.12), respectively, to obtain (6.1.9). (See also Exercise 6.1.2.)

This proves (6.1.4) when p = 2.

We now turn our attention to the case p = 2 in (6.1.4). We view (6.1.4) and (6.1.5)

as vector-valued inequalities in the spirit of Section 5.5. Define an operator T acting

on functions on Rn as follows:

T ( f )(x) = ∆ j( f )(x) j .

The inequalities (6.1.4) and (6.1.5) we wish to prove say simply that T is a bounded

operator from Lp(Rn,C) to Lp(Rn, ℓ2) and from L1(Rn,C) to L1,∞(Rn, ℓ2). We just

proved that this statement is true when p = 2, and therefore the first hypothesis of

Theorem 5.6.1 is satisfied. We observe that the operator T can be written in the form

T ( f )(x) =

∫

RnΨ2− j(x− y) f (y)dy

j

=∫

Rn

K(x− y)( f (y))dy,

where for each x ∈ Rn, K(x) is a bounded linear operator from C to ℓ2 given by

K(x)(a) = Ψ2− j(x)a j. (6.1.13)

We clearly have that ‖K(x)‖C→ℓ2 =(∑ j |Ψ2− j(x)|2

) 12 , and to be able to apply

Theorem 5.6.1 we need to know that for some constant Cn we have

∥∥K(x)∥∥

C→ℓ2 ≤Cn B |x|−n , (6.1.14)

limε↓0

∫

ε≤|y|≤1

K(y)dy =

∫ 1

0Ψ2 j(y)dy

j∈Z

, (6.1.15)

supy =0

∫

|x|≥2|y|

∥∥K(x− y)−K(x)∥∥

C→ℓ2 dx≤CnB. (6.1.16)

Of these, (6.1.14) is easily obtained using (6.1.3), (6.1.15) i.e. trivial, and so we focus

on (6.1.16). SinceΨ is a C 1 function, for |x| ≥ 2|y| we have

|Ψ2− j(x− y)−Ψ2− j(x)|≤ 2(n+1) j|∇Ψ(2 j(x−θy))| |y| for some θ ∈ [0,1],

≤ B2(n+1) j(1+2 j|x−θy|

)−(n+1)|y|

≤ B2n j(1+2 j−1|x|

)−(n+1)2 j|y| since |x−θy| ≥ 1

2|x|.

(6.1.17)

We also have that

|Ψ2− j(x− y)−Ψ2− j(x)|≤ 2n j|Ψ(2 j(x− y))|+2 jn|Ψ(2 jx)|

≤ B2n j(1+2 j|x|

)−(n+1)+B2 jn

(1+2 j−1|x|

)−(n+1)

≤ 2B2n j(1+2 j−1|x|

)−(n+1).

(6.1.18)

Taking the geometric mean of (6.1.17) and (6.1.18), we obtain for any γ ∈ [0,1]

|Ψ2− j(x− y)−Ψ2− j(x)| ≤ 21−γ B2n j(2 j|y|)γ(1+2 j−1|x|

)−(n+1). (6.1.19)

Using this estimate, when |x| ≥ 2|y|, we obtain

∥∥K(x− y)−K(x)∥∥

C→ℓ2 =

(∑j∈Z

∣∣Ψ2− j(x− y)−Ψ2− j(x)∣∣2)1/2

≤ ∑j∈Z

∣∣Ψ2− j(x− y)−Ψ2− j(x)∣∣

≤ 2B(|y| ∑

2 j< 2|x|

2(n+1) j + |y| 12 ∑

2 j≥ 2|x|

2(n+12 ) j(2 j−1|x|)−(n+1)

)

≤CnB(|y||x|−n−1 + |y| 1

2 |x|−n− 12),

where we used (6.1.19) with γ = 1 in the first sum and (6.1.19) with γ = 1/2 in

the second sum. Using this bound, we easily deduce (6.1.16) by integrating over the

region |x| ≥ 2|y|. Finally, using Theorem 5.6.1 we conclude the proofs of (6.1.4) and

(6.1.5), which establishes one direction of the theorem.

We now turn to the converse direction. Let ∆ ∗j be the adjoint operator of ∆ j

given by ∆ ∗j f = f Ψ2− j . Let f be in S ′(Rn). Then the series ∑ j∈Z∆∗j ∆ j( f ) con-

verges in S ′(Rn). To see this, it suffices to show that the sequence of partial sums

uN = ∑| j|<N ∆∗j ∆ j( f ) converges in S ′. This means that if we test this sequence

against a Schwartz function g, then it is a Cauchy sequence and hence it converges as

N → ∞. But an easy argument using duality and the Cauchy–Schwarz and Holder’s

inequalities shows that for M > N we have

|〈uN ,g〉−〈uM,g〉| ≤∥∥∥(∑

j

|∆ j( f )|2) 1

2

∥∥∥Lp

∥∥∥(∑

N≤| j|≤M

|∆ j(g)|2) 1

2∥∥∥

Lp′,

and this can be made small by picking M > N ≥N0(g). Since the sequence 〈uN ,g〉 is

Cauchy, it converges to some Λ(g). Now it remains to show that the map g →Λ(g)is a tempered distribution. Obviously Λ(g) is a linear functional. Also,

|Λ(g)| ≤∥∥∥(∑

j

|∆ j( f )|2) 1

2

∥∥∥Lp

∥∥∥(∑

j

|∆ j(g)|2) 1

2∥∥∥

Lp′

≤ Cp′

∥∥∥(∑

j

|∆ j( f )|2) 1

2

∥∥∥Lp

∥∥g∥∥

Lp′ ,

and since ‖g‖Lp′ is controlled by a finite number of Schwartz seminorms of g, it

follows that Λ is in S ′. The distribution Λ is the limit of the series ∑ j∆∗j ∆ j.

Under hypothesis (6.1.6), the Fourier transform of the tempered distribution f −∑ j∈Z∆

∗j ∆ j( f ) is supported at the origin. This implies that there exists a polynomial

Q such that f −Q = ∑ j∈Z∆∗j ∆ j( f ). Now let g be a Schwartz function. We have

∣∣⟨ f −Q , g⟩∣∣ =

∣∣⟨∑j∈Z

∆ ∗j ∆ j( f ),g⟩∣∣

=∣∣∑

j∈Z

⟨∆ ∗j ∆ j( f ),g

⟩∣∣

=∣∣∑

j∈Z

⟨∆ j( f ),∆ j(g)

⟩∣∣

=

∣∣∣∣∫

Rn∑j∈Z

∆ j( f ) ∆ j(g)dx

∣∣∣∣

≤∫

Rn

(∑j∈Z

|∆ j( f )|2) 1

2(∑j∈Z

|∆ j(g)|2) 1

2dx

≤∥∥∥(∑j∈Z

|∆ j( f )|2) 1

2∥∥∥

Lp

∥∥∥(∑j∈Z

|∆ j(g)|2) 1

2∥∥∥

Lp′


≤∥∥∥(∑j∈Z

|∆ j( f )|2) 1

2∥∥∥

LpCnBmax

(p′,(p′−1)−1

)∥∥g∥∥

Lp′ , (6.1.20)

having used the definition of the adjoint (Section 2.5.2), the Cauchy–Schwarz in-

equality, Holder’s inequality, and (6.1.4). Taking the supremum over all g in Lp′

with norm at most one, we obtain that the tempered distribution f −Q is a bounded

linear functional on Lp′ . By the Riesz representation theorem, f −Q coincides with

an Lp function whose norm satisfies the estimate

∥∥ f −Q∥∥

Lp ≤CnBmax(

p,(p−1)−1)∥∥∥

(∑j∈Z

|∆ j( f )|2) 1

2∥∥∥

Lp.

We now show uniqueness. If Q1 is another polynomial, with f −Q1 ∈ Lp, then

Q−Q1 must be an Lp function; but the only polynomial that lies in Lp is the zero

polynomial. This completes the proof of the converse of the theorem under hypoth-

esis (6.1.6).

To obtain the same conclusion under the hypothesis (6.1.7) we argue in a similar

way but we leave the details as an exercise. (One may adapt the argument in the

proof of Corollary 6.1.7 to this setting.)

Remark 6.1.3. We make some observations. If Ψ is real-valued, then the operators

∆ j are self-adjoint. Indeed,

∫

Rn∆ j( f )gdx =

∫

Rnf Ψ2− j gdξ =

∫

Rnf Ψ2− j gdξ =

∫

Rnf ∆ j(g)dx .

Moreover, ifΨ is a radial function, we see that the operators ∆ j are self-transpose,

that is, they satisfy ∫

Rn∆ j( f )gdx =

∫

Rnf ∆ j(g)dx.

Assume now thatΨ is both radial and has a real-valued Fourier transform. Suppose

also thatΨ satisfies (6.1.3) and that it has mean value zero. Then the inequality

∥∥∥∑j∈Z

∆ j( f j)∥∥∥

Lp≤CnBmax

(p,(p−1)−1

)∥∥∥(∑j∈Z

| f j|2)1

2∥∥∥

Lp(6.1.21)

is true for sequences of functions f j j. To see this we use duality. Let

T ( f ) = ∆ j( f ) j .

ThenT ∗(g j j) =∑

j

∆ j(g j) .

Inequality (6.1.4) says that the operator T maps Lp(Rn,C) to Lp(Rn, ℓ2), and its dual

statement is that T ∗ maps Lp′(Rn, ℓ2) to Lp′(Rn,C). This is exactly the statement in

(6.1.21) if p is replaced by p′. Since p is any number in (1,∞), (6.1.21) is proved.

6.1.2 Vector-Valued Analogues

We now obtain a vector-valued extension of Theorem 6.1.2. We have the following.

Proposition 6.1.4. LetΨ be an integrable C 1 function on Rn with mean value zero

that satisfies (6.1.3) and let ∆ j be the Littlewood–Paley operator associated withΨ .

Then there exists a constant Cn < ∞ such that for all 1 < p,r < ∞ and all sequences

of Lp functions f j we have

∥∥∥(∑j∈Z

(∑k∈Z

|∆k( f j)|2)r

2)1

r∥∥∥

Lp(Rn)≤CnBCp,r

∥∥∥(∑j∈Z

| f j|r)1

r∥∥∥

Lp(Rn),

where Cp,r = max(p,(p− 1)−1)max(r,(r− 1)−1). Moreover, for some C′n > 0 and

all sequences of L1 functions f j we have

∥∥∥(∑j∈Z

(∑k∈Z

|∆k( f j)|2)r

2)1

r∥∥∥

L1,∞(Rn)≤C′nBmax(r,(r−1)−1)

∥∥∥(∑j∈Z

| f j|r)1

r∥∥∥

L1(Rn).

In particular,

∥∥∥(∑j∈Z

|∆ j( f j)|r)1

r∥∥∥

Lp(Rn)≤CnBCp,r

∥∥∥(∑j∈Z

| f j|r)1

r∥∥∥

Lp(Rn). (6.1.22)

Proof. We introduce Banach spaces B1 =C and B2 = ℓ2 and for f ∈ Lp(Rn) define

an operatorT ( f ) = ∆k( f )k∈Z .

In the proof of Theorem 6.1.2 we showed that T has a kernel K that satisfies con-

dition (6.1.16). Furthermore, T obviously maps Lr(Rn,C) to Lr(Rn, ℓr). Applying

Proposition 5.6.4, we obtain the first two statements of the proposition. Restricting

to k = j yields (6.1.22).

6.1.3 Lp Estimates for Square Functions Associated with Dyadic

Sums

Let us pick a Schwartz functionΨ whose Fourier transform is compactly supported

in the annulus 2−1 ≤ |ξ | ≤ 22 such that (6.1.6) is satisfied. (Clearly (6.1.6) has no

chance of being satisfied if Ψ is supported only in the annulus 1 ≤ |ξ | ≤ 2.) The

Littlewood–Paley operation f → ∆ j( f ) represents the smoothly truncated frequency

localization of a function f near the dyadic annulus |ξ | ≈ 2 j. Theorem 6.1.2 says that

the square function formed by these localizations has Lp norm comparable to that of

the original function. In other words, this square function characterizes the Lp norm

of a function. This is the main feature of Littlewood–Paley theory.

One may ask whether Theorem 6.1.2 still holds if the Littlewood–Paley operators

∆ j are replaced by their nonsmooth versions

f →(χ2 j≤|ξ |<2 j+1 f (ξ )

)∨(x). (6.1.23)

This question has a surprising answer that already signals that there may be some

fundamental differences between one-dimensional and higher-dimensional Fourier

analysis. The square function formed by the operators in (6.1.23) can be used to

characterize Lp(R) in the same way ∆ j did, but not Lp(Rn) when n > 1 and p = 2.

The problem lies in the fact that the characteristic function of the unit disk is not

an Lp multiplier on Rn when n ≥ 2 unless p = 2; see Section 5.1 in [131]. The

one-dimensional result we alluded to earlier is the following.

For j ∈ Z we introduce the one-dimensional operator

∆ #j ( f )(x) = ( f χI j

)∨(x) , (6.1.24)

where

I j = [2 j,2 j+1)∪ (−2 j+1,−2 j] ,

and ∆ #j is a version of the operator ∆ j in which the characteristic function of the set

2 j ≤ |ξ |< 2 j+1 replaces the function Ψ(2− jξ ).

Theorem 6.1.5. There exists a constant C1 such that for all 1 < p < ∞ and all f in

Lp(R) we have

∥∥ f∥∥

Lp(Rn)

C1(p+ 1p−1

)2≤

∥∥∥(∑j∈Z

|∆ #j ( f )|2

)12∥∥∥

Lp(Rn)≤C1(p+ 1

p−1)2∥∥ f

∥∥Lp(Rn)

. (6.1.25)

Proof. Pick a Schwartz function ψ on the line whose Fourier transform is supported

in the set 2−1 ≤ |ξ | ≤ 22 and is equal to 1 on the set 1 ≤ |ξ | ≤ 2. Let ∆ j be the

Littlewood–Paley operator associated with ψ . Observe that ∆ j∆#j = ∆ #

j ∆ j = ∆ #j ,

since ψ is equal to one on the support of ∆ #j ( f ) . We now use Exercise 5.6.1(a) to

obtain

∥∥∥(∑j∈Z

|∆ #j ( f )|2

)12∥∥∥

Lp=

∥∥∥(∑j∈Z

|∆ #j ∆ j( f )|2

)12∥∥∥

Lp

≤C max(p,(p−1)−1)∥∥∥(∑j∈Z

|∆ j( f )|2)1

2∥∥∥

Lp

≤CBmax(p,(p−1)−1)2∥∥ f

∥∥Lp ,

where the last inequality follows from Theorem 6.1.2. The reverse inequality for

1 < p <∞ follows just like the reverse inequality (6.1.8) of Theorem 6.1.2 by simply

replacing the ∆ j’s by the ∆ #j ’s and setting the polynomial Q equal to zero. (There is

no need to use the Riesz representation theorem here, just the fact that the Lp norm

of f can be realized as the supremum of expressions |〈 f ,g〉| where g has Lp′ norm

at most 1.)

There is a higher-dimensional version of Theorem 6.1.5 with dyadic rectan-

gles replacing the dyadic intervals. As has already been pointed out, the higher-

dimensional version with dyadic annuli replacing the dyadic intervals is false.

Let us introduce some notation. For j ∈ Z, we denote by I j the dyadic set

[2 j,2 j+1)⋃(−2 j+1,−2 j] as in the statement of Theorem 6.1.5. For j1, . . . , jn ∈ Z

define a dyadic rectangle

R j1,..., jn = I j1 ×·· ·× I jn

in Rn. Actually R j1,..., jn is not a rectangle but a union of 2n rectangles; with some

abuse of language we still call it a rectangle. For notational convenience we write

Rj = R j1,..., jn , where j = ( j1, . . . , jn) ∈ Zn.

Observe that for different j, j′ ∈ Zn the rectangles Rj and Rj′ have disjoint interiors

and that the union of all the Rj’s is equal to Rn \0. In other words, the family of

Rj’s, where j ∈ Zn, forms a tiling of Rn, which we call the dyadic decomposition of

Rn. We now introduce operators

∆ #j ( f )(x) = ( f χRj

)∨(x) , (6.1.26)

and we have the following n-dimensional extension of Theorem 6.1.5.

Theorem 6.1.6. For a Schwartz function ψ on the line with integral zero we define

the operator

∆j( f )(x) =(ψ(2− j1ξ1) · · · ψ(2− jnξn) f (ξ )

)∨(x) , (6.1.27)

where j = ( j1, . . . , jn) ∈ Zn. Then there is a dimensional constant Cn such that

∥∥∥(∑

j∈Zn

|∆j( f )|2)1

2∥∥∥

Lp(Rn)≤Cn(p+(p−1)−1)n

∥∥ f∥∥

Lp(Rn). (6.1.28)

Let ∆ #j be the operators defined in (6.1.26). Then there exists a positive constant Cn

such that for all 1 < p < ∞ and all f ∈ Lp(Rn) we have

∥∥ f∥∥

Lp(Rn)

Cn(p+ 1p−1

)2n≤

∥∥∥(∑

j∈Zn

|∆ #j ( f )|2

)12∥∥∥

Lp(Rn)≤Cn(p+ 1

p−1)2n

∥∥ f∥∥

Lp(Rn). (6.1.29)

Proof. We first prove (6.1.28). Note that if j = ( j1, . . . , jn) ∈ Zn, then the operator

∆j is equal to

∆j( f ) = ∆( j1)j1· · ·∆ ( jn)

jn( f ) ,

where the ∆( jr)jr

are one-dimensional operators given on the Fourier transform

by multiplication by ψ(2− jrξr), with the remaining variables fixed. Inequality in

(6.1.28) is a consequence of the one-dimensional case. For instance, we discuss the

case n = 2. Using Proposition 6.1.4, we obtain

∥∥∥(∑

j∈Z2

|∆j( f )|2)1

2∥∥∥

p

Lp(R2)

=

∫

R

[∫

R

(∑

j1∈Z∑

j2∈Z

|∆ (1)j1∆(2)j2( f )(x1,x2)|2

)p2

dx1

]dx2

≤Cp max(p,(p−1)−1)p

∫

R

[∫

R

(∑

j2∈Z

|∆ (2)j2( f )(x1,x2)|2

)p2

dx1

]dx2

=Cp max(p,(p−1)−1)p

∫

R

[∫

R

(∑

j2∈Z

|∆ (2)j2( f )(x1,x2)|2

)p2

dx2

]dx1

≤C2p max(p,(p−1)−1)2p

∫

R

[∫

R| f (x1,x2)|p dx2

]dx1

=C2p max(p,(p−1)−1)2p∥∥ f

∥∥p

Lp(R2),

where we also used Theorem 6.1.2 in the calculation. Higher-dimensional versions

of this estimate may easily be obtained by induction.

We now turn to the upper inequality in (6.1.29). We pick a Schwartz function ψwhose Fourier transform is supported in the union [−4,−1/2]

⋃[1/2,4] and is equal

to 1 on [−2,−1]⋃[1,2]. Then we clearly have

∆ #j = ∆ #

j ∆j ,

since ψ(2− j1ξ1) · · · ψ(2− jnξn) is equal to 1 on the rectangle Rj. We now use Exercise

5.6.1(b) and estimate (6.1.28) to obtain

∥∥∥(∑

j∈Zn

|∆ #j ( f )|2

)12∥∥∥

Lp=

∥∥∥(∑

j∈Zn

|∆ #j ∆j( f )|2

)12∥∥∥

Lp

≤C max(p,(p−1)−1)n∥∥∥(∑

j∈Zn

|∆j( f )|2)1

2∥∥∥

Lp

≤CBmax(p,(p−1)−1)2n∥∥ f

∥∥Lp .

The lower inequality in (6.1.29) for 1 < p < ∞ is proved like inequality (6.1.8) in

Theorem 6.1.2. The fundamental ingredient in the proof is that f = ∑j∈Zn ∆ #j ∆

#j ( f )

for all Schwartz functions f , where the sum is interpreted as the L2-limit of the se-

quence of partial sums. Thus the series converges in S ′, and pairing with a Schwartz

function g, we obtain the lower inequality in (6.1.29) for Schwartz functions, by

applying the steps that prove (6.1.20) (with Q = 0). To prove the lower inequality


in (6.1.29) for a general function f ∈ Lp(Rn) we approximate an Lp function by a

sequence of Schwartz functions in the Lp norm. Then both sides of the lower in-

equality in (6.1.29) for the approximating sequence converge to the corresponding

sides of the lower inequality in (6.1.29) for f ; the convergence of the sequence of

Lp norms of the square functions requires the upper inequality in (6.1.29) that was

previously established. This concludes the proof of the theorem.

Next we observe that if the Schwartz function ψ is suitably chosen, then the

reverse inequality in estimate (6.1.28) also holds. More precisely, suppose ψ(ξ ) is

an even smooth real-valued function supported in the set 910≤ |ξ | ≤ 21

10in R that

satisfies

∑j∈Z

ψ(2− jξ ) = 1, ξ ∈ R\0; (6.1.30)

then we have the following.

Corollary 6.1.7. Suppose that ψ satisfies (6.1.30) and let ∆j be as in (6.1.27). Let f

be an Lp function on Rn such that the function(∑j∈Zn |∆j( f )|2

)12 is in Lp(Rn). Then

there is a constant Cn that depends only on the dimension and ψ such that the lower

estimate ∥∥ f∥∥

Lp

Cn(p+ 1p−1

)n≤

∥∥∥(∑

j∈Zn

|∆j( f )|2)1

2∥∥∥

Lp(6.1.31)

holds.

Proof. If we had ∑ j∈Z |ψ(2− jξ )|2 = 1 instead of (6.1.30), then we could apply the

method used in the lower estimate of Theorem 6.1.2 to obtain the required conclu-

sion. In this case we provide another argument that is very similar in spirit.

We first prove (6.1.31) for Schwartz functions f . Then the series ∑ j∈Zn ∆j( f )

converges in L2 (and hence in S ′) to f . Now let g be another Schwartz function. We

express the inner product⟨

f ,g⟩

as the action of the distribution ∑j∈Zn ∆j( f ) on the

test function g:

∣∣⟨ f ,g⟩∣∣ =

∣∣∣⟨∑

j∈Zn

∆j( f ),g⟩∣∣∣

=∣∣∣ ∑

j∈Zn

⟨∆j( f ),g

⟩∣∣∣

=∣∣∣ ∑

j∈Zn∑

k=(k1,...,kn)∈Zn

∃r |kr− jr |≤1

⟨∆j( f ),∆k(g)

⟩∣∣∣

≤∫

Rn∑

j∈Zn∑

k=(k1,...,kn)∈Zn

∃r |kr− jr |≤1

∣∣∆j( f )∣∣ ∣∣∆k(g)

∣∣dx

≤ 3n

∫

Rn

(∑

j∈Zn

∣∣∆j( f )∣∣2) 1

2(∑

k∈Zn

∣∣∆k(g)∣∣2) 1

2dx

≤ 3n∥∥∥(∑

j∈Zn

∣∣∆j( f )∣∣2) 1

2∥∥∥

Lp

∥∥∥(∑

k∈Zn

∣∣∆k(g)∣∣2) 1

2∥∥∥

Lp′

≤C−1n max

(p′,(p′−1)−1

)n∥∥g∥∥

Lp′

∥∥∥(∑

j∈Zn

∣∣∆j( f )∣∣2) 1

2∥∥∥

Lp,

where we used the fact that ∆j( f ) and ∆k(g) are orthogonal operators unless every

coordinate of k is within 1 unit of the corresponding coordinate of j; this is an easy

consequence of the support properties of ψ . We now take the supremum over all g

in Lp′ with norm at most 1, to obtain (6.1.31) for Schwartz functions f .

To extend this estimate to general Lp functions f , we use the density argument

described in the last paragraph in the proof of Theorem 6.1.6.

6.1.4 Lack of Orthogonality on Lp

We discuss two examples indicating why (6.1.1) cannot hold if the exponent 2 is

replaced by some other exponent q = 2. More precisely, we show that if the functions

f j have Fourier transforms supported in disjoint sets, then the inequality

∥∥∥∑j

f j

∥∥∥p

Lp≤Cp∑

j

∥∥ f j

∥∥p

Lp (6.1.32)

cannot hold if p > 2, and similarly, the inequality

∑j

∥∥ f j

∥∥p

Lp ≤Cp

∥∥∥∑j

f j

∥∥∥p

Lp(6.1.33)

cannot hold if p < 2. In both (6.1.32) and (6.1.33) the constants Cp are supposed to

be independent of the functions f j.

Example 6.1.8. Pick a Schwartz function ζ whose Fourier transform is positive and

supported in the interval |ξ | ≤ 1/4. Let N be a large integer and let

f j(x) = e2πi jxζ (x).

Then

f j(ξ ) = ζ (ξ − j)

and the f j’s have disjoint Fourier transforms. We obviously have

N

∑j=0

∥∥ f j

∥∥p

Lp = (N +1)∥∥ζ

∥∥p

Lp .

On the other hand, we have the estimate

∥∥∥N

∑j=0

f j

∥∥∥p

Lp=

∫

R

∣∣ e2πi(N+1)x−1e2πix−1

∣∣p|ζ (x)|p dx

≥ c

∫

|x|< 110 (N+1)−1

(N+1)p|x|p|x|p |ζ (x)|p dx

=Cζ (N +1)p−1 ,

since ζ does not vanish in a neighborhood of zero. We conclude that (6.1.32) cannot

hold for this choice of f j’s for p > 2.

Example 6.1.9. We now indicate why (6.1.33) cannot hold for p < 2. We pick a

smooth functionΨ on the line whose Fourier transform Ψ is supported in[

78, 17

8

],

is nonnegative, is equal to 1 on[

98, 15

8

], and satisfies

∑j∈Z

Ψ(2− jξ )2 = 1, ξ > 0.

Extend Ψ to be an even function on the whole line and let ∆ j be the Littlewood–

Paley operator associated withΨ . Also pick a nonzero Schwartz function ϕ on the

real line whose Fourier transform is nonnegative and supported in the set[

118, 13

8

].

Fix N a large positive integer and let

f j(x) = e2πi 128 2 jxϕ(x), (6.1.34)

for j = 1,2, . . . ,N. Then the function f j(ξ ) = ϕ(ξ − 128

2 j) is supported in the set[118+ 12

82 j, 13

8+ 12

82 j], which is contained in

[982 j, 15

82 j]

for j ≥ 3. In other words,

Ψ(2− jξ ) is equal to 1 on the support of f j. This implies that

∆ j( f j) = f j for j ≥ 3.

This observation combined with (6.1.21) gives for N ≥ 3,

∥∥∥N

∑j=3

f j

∥∥∥Lp

=∥∥∥

N

∑j=3

∆ j( f j)∥∥∥

Lp≤Cp

∥∥∥( N

∑j=3

| f j|2)1

2∥∥∥

Lp=Cp

∥∥ϕ∥∥

Lp(N−2)12 ,

where 1 < p < ∞. On the other hand, (6.1.34) trivially yields that

( N

∑j=3

∥∥ f j

∥∥p

Lp

) 1p=

∥∥ϕ∥∥

Lp(N−2)1p .

Letting N → ∞ we see that (6.1.33) cannot hold for p < 2 even when the f j’s have

Fourier transforms supported in disjoint sets.

Example 6.1.10. A similar idea illustrates the necessity of the ℓ2 norm in (6.1.4). To

see this, letΨ and ∆ j be as in Example 6.1.9. Let us fix 1 < p < ∞ and q < 2. We

show that the inequality

∥∥∥(∑j∈Z

|∆ j( f )|q)1

q∥∥∥

Lp≤Cp,q

∥∥ f∥∥

Lp (6.1.35)

cannot hold. Take f = ∑Nj=3 f j, where the f j are as in (6.1.34) and N ≥ 3. Then

the left-hand side of (6.1.35) is bounded from below by ‖ϕ‖Lp(N−2)1/q, while the

right-hand side is bounded above by ‖ϕ‖Lp(N− 2)1/2. Letting N → ∞, we deduce

that (6.1.35) is impossible when q < 2.

Example 6.1.11. For 1 < p < ∞ and 2 < q < ∞, the inequality

∥∥g∥∥

Lp ≤Cp,q

∥∥∥(∑j∈Z

|∆ j(g)|q)1

q∥∥∥

Lp(6.1.36)

cannot hold even under assumption (6.1.6) on Ψ . Let ∆ j be as in Example 6.1.9.

Let us suppose that (6.1.36) did hold for some q > 2 for these ∆ j’s. Then the self-

adjointness of the ∆ j’s and duality would give

∥∥∥(∑k∈Z

|∆k(g)|q′) 1

q′∥∥∥

Lp′

= sup∥∥‖hkk‖ℓq∥∥

Lp≤1

∣∣∣∣∫

R∑k∈Z

∆k(g)hk dx

∣∣∣∣

≤∥∥g

∥∥Lp′ sup∥∥‖hkk‖ℓq

∥∥Lp≤1

∥∥∥∑k∈Z

∆k(hk)∥∥∥

Lp

≤C∥∥g


∥∥Lp≤1

∥∥∥(∑j∈Z

∣∣∣∆ j

(∑k∈Z

∆k(hk))∣∣∣

q)1q∥∥∥

Lpby (6.1.36)

≤C′∥∥g


∥∥Lp≤1

1

∑l=−1

∥∥∥(∑j∈Z

|∆ j∆ j+l(h j)|q)1

q∥∥∥

Lp

≤C′′∥∥g


∥∥Lp≤1

∥∥∥(∑j∈Z

|h j|q)1

q∥∥∥

Lp=C′′

∥∥g∥∥

Lp′ ,

where the next-to-last inequality follows from (6.1.22) applied twice, while the one

before that follows from support considerations. But since q′ < 2, this exactly proves

(6.1.35), previously shown to be false, a contradiction.

We conclude that if both assertions (6.1.4) and (6.1.8) of Theorem 6.1.2 were to

hold, then the ℓ2 norm inside the Lp norm could not be replaced by an ℓq norm for

some q = 2. Exercise 6.1.6 indicates the crucial use of the fact that ℓ2 is a Hilbert

space in the converse inequality (6.1.8) of Theorem 6.1.2.

Exercises

6.1.1. Construct a Schwartz function Ψ that satisfies ∑ j∈Z |Ψ(2− jξ )|2 = 1 for all

ξ ∈ Rn \0 and whose Fourier transform is supported in the annulus 67≤ |ξ | ≤ 2

and is equal to 1 on the annulus 1≤ |ξ | ≤ 147

.[Hint: Set Ψ(ξ ) = η(ξ )

(∑k∈Z |η(2−kξ )|2

)−1/2for a suitable η ∈ C ∞

0 (Rn) .]

6.1.2. Suppose that Ψ is an integrable function on Rn that satisfies |Ψ(ξ )| ≤B min(|ξ |ε , |ξ |−ε ′) for some ε ′,ε > 0. Show that for some constant Cε ,ε ′ < ∞ we

have

supξ∈Rn

(∫ ∞

0|Ψ(tξ )|2 dt

t

) 12

+ supξ∈Rn

(∑j∈Z

|Ψ(2− jξ )|2) 1

2 ≤Cε ,ε ′ B .

6.1.3. LetΨ be an integrable function on Rn with mean value zero that satisfies

|Ψ(x)| ≤ B(1+ |x|)−n−ε ,∫

Rn|Ψ(x− y)−Ψ(x)|dx≤ B |y|ε ′ ,

for some B,ε ′,ε > 0 and for all y = 0.

(a) Prove that |Ψ(ξ )| ≤ cn,ε ,ε ′ B min(|ξ |min( ε2 ,1), |ξ |−ε) for some constant cn,ε ,ε ′ and

conclude that (6.1.4) holds for p = 2.

(b) Deduce the validity of (6.1.4) and (6.1.5).

(c) If ε < 1 and the assumption |Ψ(x)| ≤ B(1+ |x|)−n−ε is weakened to |Ψ(x)| ≤B |x|−n−ε for all x ∈ Rn, then show that |Ψ(ξ )| ≤ cn,ε ,ε ′ B min(|ξ | ε2 , |ξ |−ε) and thus

(6.1.4) and (6.1.5) are valid.[Hint: Part (a): Make use of the identity

Ψ(ξ ) =∫

Rne−2πix·ξΨ(x)dx =−

∫

Rne−2πix·ξΨ(x− y)dx ,

where y= 12

ξ|ξ |2 when |ξ | ≥ 1. For |ξ | ≤ 1 use the mean value property ofΨ to write

Ψ(ξ ) =∫|x|≤1Ψ(x)(e−2πix·ξ −1)dx and split the integral in the regions |x| ≤ 1 and

|x| ≥ 1. Part (b): If K is defined by (6.1.13), then control the ℓ2(Z) norm by the ℓ1(Z)norm to prove (6.1.16). Then split the sum ∑ j∈Z

∫|x|≥2|y|

∣∣Ψ2− j(x− y)−Ψ2− j(x)∣∣dx

into the parts ∑2 j≤|y|−1 and ∑2 j>|y|−1 . Part (c): Notice that when ε < 1, we have

|∫|x|≤1Ψ(x)(e−2πix·ξ −1)dx| ≤Cn B |ξ | ε2 .]

6.1.4. LetΨ be an integrable function on Rn with mean value zero that satisfies

|Ψ(x)| ≤ B(1+ |x|)−n−ε ,∫

Rn|Ψ(x− y)−Ψ(x)|dx≤ B|y|ε ′ ,

for some B,ε ′,ε > 0 and for all y = 0. Let Ψt(x) = t−nΨ(x/t). (a) Prove that there

are constants cn,c′n such that

(∫ ∞

0|Ψt(x)|2

dt

tdx

) 12

≤ cn B |x|−n ,

supy∈Rn\0

∫

|x|≥2|y|

(∫ ∞

0|Ψt(x− y)−Ψt(x)|2

dt

t

) 12

dx≤ c′n B .

(b) Show that there exist constants Cn,C′n such that for all 1 < p < ∞ and for all

f ∈ Lp(Rn) we have

∥∥∥(∫ ∞

0| f ∗Ψt |2

dt

t

)12∥∥∥

Lp(Rn)≤CnBmax(p,(p−1)−1)

∥∥ f∥∥

Lp(Rn)

and also for all f ∈ L1(Rn) we have

∥∥∥(∫ ∞

0| f ∗Ψt |2

dt

t

)12∥∥∥

L1,∞(Rn)≤C′nB

∥∥ f∥∥

L1(Rn).

(c) Under the additional hypothesis that 0 <∫ ∞

0 |Ψ(tξ )|2 dtt= c0 for all ξ ∈Rn \0,

prove that for all f ∈ Lp(Rn) we have

∥∥ f∥∥

Lp(Rn)≤C′′n B max(p,(p−1)−1)

∥∥∥(∫ ∞

0| f ∗Ψt |2

dt

t

)12∥∥∥

Lp(Rn)

[Hint: Part (a): Use the Cauchy-Schwarz inequality to obtain

∫

|x|≥2|y|

(∫ ∞

0|Ψt(x− y)−Ψt(x)|2

dt

t

) 12

dx

≤ cn|y|−ε2

(∫

|x|≥2|y||x|n+ε

∫ ∞

0|Ψt(x− y)−Ψt(x)|2

dt

tdx

) 12

,

and split the integral on the right into the regions t ≤ |y| and t > |y|. In the second

region use thatΨ is bounded to replace the square by the first power. Part (b): Use

Exercise 6.1.2 and part (a) of Exercise 6.1.3 and to deduce the inequality when p= 2.

Then apply Theorem 5.6.1. Part (c): Prove the inequality first for f ∈S (Rn) using

duality.]

6.1.5. Prove the following generalization of Theorem 6.1.2. Let A > 0. Suppose that

K j j∈Z is a sequence of locally integrable functions on Rn \0 that satisfies

supx =0

|x|n(∑j∈Z

|K j(x)|2) 1

2 ≤ A ,

supy∈Rn\0

∫

|x|≥2|y|

(∑j∈Z

|K j(x− y)−K j(x)|2)1

2dx≤ A < ∞ ,

and for each j ∈ Z there is a number L j such that

limεk↓0

∫

εk≤|y|≤1K j(y)dy = L j .

If the K j coincide with tempered distributions Wj that satisfy

∑j∈Z

|Wj(ξ )|2 ≤ B2 ,

then the operator

f →(∑j∈Z

|K j ∗ f |2)1

2

maps Lp(Rn) to itself and is weak type (1,1) norms at most multiples of A+B.

6.1.6. Suppose that H is a Hilbert space with inner product 〈 · , · 〉H . Let A > 0 and

1 < p < ∞. Suppose that an operator T from L2(Rn)→ L2(Rn,H ) is a multiple of

an isometry, that is, ∥∥T (g)∥∥

L2(Rn,H )= A

∥∥g∥∥

L2(Rn)

for all g ∈ L2(Rn,H ). Then the inequality ‖T ( f )‖Lp(Rn,H ) ≤ Cp‖ f‖Lp(Rn) for all

f ∈S (Rn) implies

∥∥ f∥∥

Lp′ (Rn)≤Cp′A

−2∥∥T ( f )

∥∥Lp′ (Rn,H )

for all in f ∈S (Rn).[Hint: Use the inner product structure and polarization to obtain

A2

∣∣∣∣∫

Rnf (x)g(x)dx

∣∣∣∣=∣∣∣∣∫

Rn

⟨T ( f )(x),T (g)(x)

⟩H

dx

∣∣∣∣

and then argue as in the proof of inequality (6.1.8).]

6.1.7. Suppose that m j j∈Z is a sequence of bounded functions supported in the

intervals [2 j,2 j+1]. Let Tj( f ) = ( f m j)∨ be the corresponding multiplier operators.

Assume that for all sequences of functions f j j the vector-valued inequality

∥∥∥(∑

j

|Tj( f j)|2)1

2∥∥∥

Lp≤ Ap

∥∥∥(∑

j

| f j|2)1

2∥∥∥

Lp

is valid for some 1 0 such that for all finite subsets S

of Z we have ∥∥∥∑j∈S

m j

∥∥∥Mp

≤Cp Ap.

[Hint: Use that

⟨∑ j∈S Tj( f ),g

⟩= ∑ j∈S

⟨∆ #

j Tj( f ),∆ #j (g)

⟩.]

6.2 Two Multiplier Theorems 437

6.1.8. Let m be a bounded function on Rn that is supported in the annulus

1 ≤ |ξ | ≤ 2 and define Tj( f ) =(

f (ξ )m(2− jξ ))∨

. Suppose that the square func-

tion f →(∑ j∈Z |Tj( f )|2

)1/2is bounded on Lp(Rn) for some 1 < p < ∞. Show that

for every finite subset S of the integers we have

∥∥∥∑j∈S

Tj( f )∥∥∥

Lp(Rn)≤Cp,n

∥∥ f∥∥

Lp(Rn)

for some constant Cp,n independent of S.

6.1.9. Fix a nonzero Schwartz function h on the line whose Fourier transform is

supported in the interval[− 1

8, 1

8

]. For a j a sequence of numbers, set

f (x) =∞

∑j=1

a je2πi2 jxh(x) .

Prove that for all 1 < p < ∞ there exists a constant Cp such that

‖ f‖Lp(R) ≤Cp

(∑

j

|a j|2) 1

2 ‖h‖Lp .

[Hint: Write f = ∑∞j=1∆ j(a je

2πi2 j(·)h), where ∆ j is given by convolution with ϕ2− j

for some ϕ whose Fourier transform is supported in the interval[

68, 10

8

]and is equal

to 1 on[

78, 9

8

]. Then use (6.1.21).

]

6.1.10. Let Ψ be a Schwartz function whose Fourier transform is supported in the

annulus 12≤ |ξ | ≤ 2 and that satisfies (6.1.7). Define a Schwartz function Φ by

setting

Φ(ξ ) =

∑ j≤0Ψ(2− jξ ) when ξ = 0,

1 when ξ = 0.

Let S0 be the operator given by convolution with Φ . Let 1 < p <∞ and f ∈ Lp(Rn).Show that

∥∥ f∥∥

Lp ≈∥∥S0( f )

∥∥Lp +

∥∥∥( ∞

∑j=1

|∆ j( f )|2) 1

2∥∥∥

Lp.

[Hint: Use Theorem 6.1.2 together with the identity S0 +∑∞j=1∆ j = I.

]

6.2 Two Multiplier Theorems

We now return to the spaces Mp introduced in Section 2.5. We seek sufficient con-

ditions on L∞ functions defined on Rn to be elements of Mp. In this section we are

concerned with two fundamental theorems that provide such sufficient conditions.


These are the Marcinkiewicz and the Hormander–Mihlin multiplier theorems. Both

multiplier theorems are consequences of the Littlewood–Paley theory discussed in

the previous section.

Using the dyadic decomposition of Rn, we can write any L∞ function m as the

sum

m = ∑j∈Zn

mχRja.e.,

where j = ( j1, . . . , jn), Rj = I j1 ×·· ·× I jn , and Ik = [2k,2k+1)⋃(−2k+1,−2k]. For j

in Zn we set mj =mχRj. A consequence of the ideas developed so far is the following

characterization of Mp(Rn) in terms of a vector-valued inequality.

Proposition 6.2.1. Let m ∈ L∞(Rn) and let mj = mχRj. Then m lies in Mp(R

n), that

is, for some cp we have

∥∥( f m)∨∥∥

Lp ≤ cp

∥∥ f∥∥

Lp , f ∈ Lp(Rn),

if and only if for some Cp > 0 we have

∥∥∥(∑

j∈Zn

|( fjmj)∨|2

)12∥∥∥

Lp≤Cp

∥∥∥(∑

j∈Zn

| fj|2)1

2∥∥∥

Lp(6.2.1)

for all sequences of functions fj in Lp(Rn).

Proof. Suppose that m ∈Mp(Rn). Exercise 5.6.1 gives the first inequality below

∥∥∥(∑

j∈Zn

|(χRjm fj)

∨|2)1

2∥∥∥

Lp≤Cp

∥∥∥(∑

j∈Zn

|(m fj)∨|2

)12∥∥∥

Lp≤Cp

∥∥∥(∑

j∈Zn

| fj|2)1

2∥∥∥

Lp,

while the second inequality follows from Theorem 5.5.1. (Observe that when p = q

in Theorem 5.5.1, then Cp,q = 1.) Conversely, suppose that (6.2.1) holds for all se-

quences of functions fj. Fix a function f and apply (6.2.1) to the sequence ( f χRj)∨,

where Rj is the dyadic rectangle indexed by j = ( j1, . . . , jn) ∈ Zn. We obtain

∥∥∥(∑

j∈Zn

|( f mχRj)∨|2

)12∥∥∥

Lp≤Cp

∥∥∥(∑

j∈Zn

|( f χRj)∨|2

)12∥∥∥

Lp.

Using Theorem 6.1.6, we obtain that the previous inequality is equivalent to the

inequality ∥∥( f m)∨∥∥

Lp ≤ cp

∥∥ f∥∥

Lp ,

which implies that m ∈Mp(Rn).

6.2.1 The Marcinkiewicz Multiplier Theorem on R

Proposition 6.2.1 suggests that the behavior of m on each dyadic rectangle R j should

play a crucial role in determining whether m is an Lp multiplier. The Marcinkiewicz

multiplier theorem provides such sufficient conditions on m restricted to any dyadic

rectangle R j. Before stating this theorem, we illustrate its main idea via the follow-

ing example. Suppose that m is a bounded function that vanishes near −∞, that is

differentiable at every point, and whose derivative is integrable. Then we may write

m(ξ ) =∫ ξ

−∞m′(t)dt =

∫ +∞

−∞χ[t,∞)(ξ )m

′(t)dt ,

from which it follows that for a Schwartz function f we have

( f m)∨ =∫

R( f χ[t,∞))

∨m′(t)dt.

Since the operators f → ( f χ[t,∞))∨ map Lp(R) to itself independently of t, it follows

that ∥∥( f m)∨∥∥

Lp ≤Cp

∥∥m′∥∥

L1

∥∥ f∥∥

Lp ,

thus yielding that m is in Mp(R). The next multiplier theorem is an improvement

of this result and is based on the Littlewood–Paley theorem. We begin with the one-

dimensional case, which already captures the main ideas.

Theorem 6.2.2. (Marcinkiewicz multiplier theorem) Let m : R→ R be a bounded

function that is C 1 in every dyadic set (2 j,2 j+1)⋃(−2 j+1,−2 j) for j ∈ Z. Assume

that the derivative m′ of m satisfies

supj

[∫ −2 j

−2 j+1|m′(ξ )|dξ +

∫ 2 j+1

2 j|m′(ξ )|dξ

]≤ A < ∞ . (6.2.2)

Then for all 1 0 we have

∥∥m∥∥

Mp(R)≤C max

(p,(p−1)−1

)6(∥∥m∥∥

L∞+A

). (6.2.3)

Proof. Since the function m has an integrable derivative on (2 j,2 j+1), it has

bounded variation in this interval and hence it is a difference of two increasing

functions. Therefore, m has left and right limits at the points 2 j and 2 j+1, and by

redefining m at these points we may assume that m is right continuous at the points

2 j and left continuous at the points −2 j.

Set I j = [2 j,2 j+1)∪ (−2 j+1,−2 j] and I+j = [2 j,2 j+1) whenever j ∈ Z. Given

an interval I in R, we introduce an operator ∆I defined by ∆I( f ) = ( f χI)∨. With

this notation ∆I+j( f ) is “half” of the operator ∆ #

j introduced in the previous section.

Given m as in the statement of the theorem, we write m(ξ ) = m+(ξ ) + m−(ξ ),where m+(ξ ) = m(ξ )χξ≥0 and m−(ξ ) = m(ξ )χξ<0. We show that both m+ and m−

are Lp multipliers. Since m′ is integrable over all intervals of the form [2 j,ξ ] when

2 j ≤ ξ < 2 j+1, the fundamental theorem of calculus gives

m(ξ ) = m(2 j)+

∫ ξ

2 jm′(t)dt, for 2 j ≤ ξ < 2 j+1,

from which it follows that for a Schwartz function f on the real line we have

m(ξ ) f (ξ )χI+j(ξ ) = m(2 j) f (ξ )χI+j

(ξ )+∫ 2 j+1

2 jf (ξ )χ[t,∞)(ξ )χI+j

(ξ )m′(t)dt .

We therefore obtain the identity

( f χI jm+)

∨ = ( f mχI+j)∨ = m(2 j)∆I+j

( f )+

∫ 2 j+1

2 j∆[t,∞)∆I+j

( f )m′(t)dt ,

which implies that

|( f χI jm+)

∨| ≤ ‖m‖L∞ |∆I+j( f )|+A

12

(∫ 2 j+1

2 j

∣∣∆[t,∞)∆I+j( f )

∣∣2 |m′(t)|dt

)12

,

using the hypothesis (6.2.2). Taking ℓ2(Z) norms we obtain

(∑j∈Z

|( f χI jm+)

∨|2)1

2 ≤ ‖m‖L∞

(∑j∈Z

|∆I+j( f )|2

)12

+A12

(∫ ∞

0

∣∣∆[t,∞)∆#[log2 t]( f )

∣∣2 |m′(t)|dt

)12

.

Exercise 5.6.2 gives

A12

∥∥∥(∫ ∞

0

∣∣∆[t,∞)∆#[log2 t]( f )

∣∣2|m′(t)|dt)1

2∥∥∥

Lp

≤C max(p,(p−1)−1)A12

∥∥∥(∫ ∞

0

∣∣∆ #[log2 t]( f )

∣∣2|m′(t)|dt)1

2∥∥∥

Lp,

while the hypothesis on m′ implies the inequality

∥∥∥(∑j∈Z

∣∣∆I+j( f )

∣∣2∫

I+j

|m′(t)|dt)1

2∥∥∥

Lp≤ A

12

∥∥∥(∑

j

|∆I+j( f )|2

)12∥∥∥

Lp.

Using Theorem 6.1.5 we obtain that

∥∥∥(∑

j

|∆I+j( f )|2

)12∥∥∥

Lp≤C′max(p,(p−1)−1)2

∥∥( f χ(0,∞))∨∥∥

Lp ,

and the latter is at most a constant multiple of max(p,(p− 1)−1)3∥∥ f

∥∥Lp . Putting

things together we deduce that

∥∥∥(∑

j

|( f χI jm+)

∨|2)1

2∥∥∥

Lp≤C′′max(p,(p−1)−1)4

(A+‖m‖L∞

)∥∥ f∥∥

Lp , (6.2.4)


∥∥( f m+)∨∥∥

Lp ≤C max(p,(p−1)−1)6(A+‖m‖L∞

)∥∥ f∥∥

Lp ,

using the lower estimate of Theorem 6.1.5. This proves (6.2.3) for m+. A similar

argument also works for m−, and this concludes the proof by summing the corre-

sponding estimates for m+ and m−.

We remark that the same proof applies under the more general assumption that

m is a function of bounded variation on every interval [2 j,2 j+1] and [−2 j+1,−2 j].In this case the measure |m′(t)|dt should be replaced by the total variation |dm(t)|of the Lebesgue–Stieltjes measure dm(t).

Example 6.2.3. Any bounded function that is constant on dyadic intervals is an Lp

multiplier. Also, the function

m(ξ ) = |ξ |2−[log2 |ξ |]

is an Lp multiplier on R for 1 < p < ∞.

6.2.2 The Marcinkiewicz Multiplier Theorem on Rn

We now extend this theorem on Rn. As usual we denote the coordinates of a point

ξ ∈ Rn by (ξ1, . . . ,ξn). We recall the notation I j = (−2 j+1,−2 j]⋃[2 j,2 j+1) and

Rj = I j1 ×·· ·× I jn whenever j = ( j1, . . . , jn) ∈ Zn.

Theorem 6.2.4. Let m be a bounded function on Rn such that for all α =(α1, . . . ,αn)with |α1|, . . . , |αn| ≤ 1 the derivatives ∂αm are continuous up to the boundary of Rj

for all j ∈ Zn. Assume that there is a constant A < ∞ such that for all partitions

s1, . . . ,sk∪r1, . . . ,rℓ= 1,2, . . . ,n with n = k+ ℓ and all ξ ∈ Rj we have

supξr1∈I jr1

· · · supξrℓ∈I jrℓ

∫

I js1

· · ·∫

I jsk

∣∣(∂s1· · ·∂sk

m)(ξ1, . . . ,ξn)∣∣dξsk

· · ·dξs1≤ A (6.2.5)

for all j = ( j1, . . . , jn) ∈ Zn. Then m is in Mp(Rn) whenever 1 < p < ∞ and there is

a constant Cn < ∞ such that

‖m‖Mp(Rn) ≤Cn

(A+‖m‖L∞

)max

(p,(p−1)−1

)6n. (6.2.6)

Proof. We prove this theorem only in dimension n = 2, since the general case

presents no substantial differences but only some notational inconvenience. We de-

compose the given function m as

m(ξ ) = m++(ξ )+m−+(ξ )+m+−(ξ )+m−−(ξ ) ,

where each of the last four terms is supported in one of the four quadrants. For

instance, the function m+−(ξ1,ξ2) is supported in the quadrant ξ1 ≥ 0 and ξ2 < 0.

As in the one-dimensional case, we work with each of these pieces separately. By

symmetry we choose to work with m++ in the following argument.

Using the fundamental theorem of calculus, we obtain the following simple iden-

tity, valid for 2 j1 ≤ ξ1 < 2 j1+1 and 2 j2 ≤ ξ2 < 2 j2+1:

m(ξ1,ξ2) = m(2 j1 ,2 j2)+∫ ξ1

2 j1

(∂1m)(t1,2j2)dt1

+∫ ξ2

2 j2

(∂2m)(2 j1 , t2)dt2

+∫ ξ1

2 j1

∫ ξ2

2 j2

(∂1∂2m)(t1, t2)dt2 dt1 .

(6.2.7)

We introduce operators ∆(r)I , r ∈ 1,2, acting in the rth variable (with the other

variable remaining fixed) given by multiplication on the Fourier transform side by

the characteristic function of the interval I. Likewise, we introduce operators ∆#(r)j ,

r ∈ 1,2 (also acting in the rth variable), given by multiplication on the Fourier

transform side by the characteristic function of the set (−2 j+1,−2 j]⋃[2 j,2 j+1). For

notational convenience, for a given Schwartz function f we write

f++ =(

f χ(0,∞)2

)∨,

and likewise we define f+−, f−+, and f−−.

Multiplying both sides of (6.2.7) by the function f χRjχ(0,∞)2 and taking inverse

Fourier transforms yields

( f χRjm++)

∨ = m(2 j1 ,2 j2)∆#(1)j1

∆#(2)j2

( f++)

+∫ 2 j1+1

2 j1

∆#(2)j2

∆(1)[t1,∞)

∆#(1)j1

( f++)(∂1m)(t1,2j2)dt1

+∫ 2 j2+1

2 j2

∆#(1)j1

∆(2)[t2,∞)

∆#(2)j2

( f++)(∂2m)(2 j1 , t2)dt2

+

∫ 2 j1+1

2 j1

∫ 2 j2+1

2 j2

∆(1)[t1,∞)

∆#(1)j1

∆(2)[t2,∞)

∆#(2)j2

( f++)(∂1∂2m)(t1, t2)dt2 dt1 .

(6.2.8)

We apply the Cauchy–Schwarz inequality in the last three terms of (6.2.8) with re-

spect to the measures |(∂1m)(t1,2j2)|dt1, |(∂2m)(2 j1 , t2)|dt2, |(∂1∂2m)(t1, t2)|dt2dt1

and we use hypothesis (6.2.5) to deduce


∣∣( f χRjm++)

∨∣∣ ≤ ‖m‖L∞∣∣∆ #(1)

j1∆

#(2)j2

( f++)∣∣

+ A12

(∫ 2 j1+1

2 j1

∣∣∆ #(2)j2

∆(1)[t1,∞)

∆#(1)j1

( f++)∣∣2 |(∂1m)(t1,2

j2)|dt1

) 12

+ A12

(∫ 2 j2+1

2 j2

∣∣∆ #(1)j1

∆(2)[t2,∞)

∆#(2)j2

( f++)∣∣2 |(∂2m)(2 j1 , t2)|dt2

) 12

+ A12

(∫ 2 j1+1

2 j1

∫ 2 j2+1

2 j2

∣∣∆ (1)[t1,∞)

∆#(1)j1

∆(2)[t2,∞)

∆#(2)j2

( f++)∣∣2 |(∂1∂2m)(t1, t2)|dt2 dt1

) 12

.

Both sides of the preceding inequality are sequences indexed by j ∈ Z2. We apply

ℓ2(Z2) norms and use Minkowski’s inequality to deduce the pointwise estimate

(∑

j∈Z2

∣∣( f χRjm++)

∨∣∣2)1

2 ≤ ‖m‖L∞

(∑

j∈Z2

∣∣∆ #j ( f++)

∣∣2)1

2

+A12

(∫ ∞

0

∫ ∞

0

∣∣∆ (1)[t1,∞)

∆#(2)[log2 t2]

∆#(1)[log2 t1]

( f++)∣∣2 ∣∣(∂1m)(t1,2

[log2 t2])∣∣dt1dν(t2)

)12

+A12

(∫ ∞

0

∫ ∞

0

∣∣∆ (2)[t2,∞)

∆#(1)[log2 t1]

∆#(2)[log2 t2]

( f++)∣∣2 ∣∣(∂2m)(2[log2 t1], t2)

∣∣dν(t1)dt2

)12

+A12

(∫ ∞

0

∫ ∞

0

∣∣∆ (1)[t1,∞)

∆(2)[t2,∞)

∆#(1)[log2 t1]

∆#(2)[log2 t2]

( f++)∣∣2∣∣(∂1∂2m)(t1, t2)

∣∣dt1dt2

)12

,

where ν is the counting measure ∑ j∈Z δ2 j defined by ν(A) = # j ∈ Z : 2 j ∈ Afor subsets A of (0,∞). We now take Lp(R2) norms and we estimate separately the

contribution of each of the four terms on the right side. Using Exercise 5.6.2 we

obtain

∥∥∥∥(∑

j∈Z2

∣∣( f χRjm++)

∨∣∣2)1

2

∥∥∥∥Lp

≤ ‖m‖L∞

∥∥∥∥(∑

j∈Z2

∣∣∆ #j ( f++)

∣∣2)1

2

∥∥∥∥Lp

+ C2 A12 max

(p,(p−1)−1

)2

×∥∥∥∥

(∫ ∞

0

∫ ∞

0

∣∣∆ #(2)[log2 t2]

∆#(1)[log2 t1]

( f++)∣∣2 ∣∣(∂1m)(t1,2

[log2 t2])∣∣dt1 dν(t2)

)12∥∥∥∥

Lp

+

∥∥∥∥(∫ ∞

0

∫ ∞

0

∣∣∆ #(1)[log2 t1]

∆#(2)[log2 t2]

( f++)∣∣2 ∣∣(∂2m)(2[log2 t1], t2)

∣∣dν(t1)dt2

)12∥∥∥∥

Lp

+

∥∥∥∥(∫ ∞

0

∫ ∞

0

∣∣∆ #(1)[log2 t1]

∆#(2)[log2 t2]

( f++)∣∣2∣∣(∂1∂2m)(t1, t2)

∣∣dt1dt2

)12∥∥∥∥

Lp

.


But the functions (t1, t2) → ∆#(1)[log2 t1]

∆#(2)[log2 t2]

( f++) are constant on products of inter-

vals of the form [2 j1 ,2 j1+1)× [2 j2 ,2 j2+1); hence using hypothesis (6.2.5) again we

deduce the estimate

∥∥∥∥(∑

j∈Z2

∣∣( f χRjm++)

∨∣∣2)1

2

∥∥∥∥Lp(R2)

≤C2

(‖m‖L∞ +A

)max

(p,(p−1)−1

)2

∥∥∥∥(∑

j∈Z2

∣∣∆ #j ( f++)

∣∣2)1

2

∥∥∥∥Lp(R2)

≤C2

(‖m‖L∞ +A

)max

(p,(p−1)−1

)6∥∥( f χ(0,∞)2)∨∥∥

Lp(R2)

≤C2

(‖m‖L∞ +A

)max

(p,(p−1)−1

)8∥∥ f∥∥

Lp(R2),

where the penultimate estimate follows from Theorem 6.1.6 and the last estimate

by the boundedness of the Hilbert transform (Theorem 5.1.7). We now appeal

to inequality (6.1.29) which yields the required estimate for the Lp(R2) norm of

( f m++)∨. A similar argument also works for the remaining parts of m+−, m−+,

m−−, and summing concludes the proof of (6.2.6).

The analogous estimate on Rn is

∥∥∥∥(∑

j∈Zn

∣∣( f χRjm+···+)

∨∣∣2)1

2

∥∥∥∥Lp(Rn)

≤Cn

(‖m‖L∞ +A

)max

(p,(p−1)−1

)4n∥∥ f∥∥

Lp(Rn)

which is obtained in a similar fashion. Using (6.1.29), this implies that

∥∥( f m+···+)∨∥∥

Lp(Rn)≤Cn

(‖m‖L∞ +A

)max

(p,(p−1)−1

)6n∥∥ f∥∥

Lp(Rn).

A similar inequality holds when some (or all) +’s are replaced by −’s.

We now give a condition that implies (6.2.5) and is well suited for a variety of

applications.

Corollary 6.2.5. Let m be a bounded C n function defined away from the coordinate

axes on Rn. Assume that for all k ∈ 1, . . . ,n, all distinct j1, . . . , jk ∈ 1,2, . . . ,n,and all ξr ∈ R\0 for r /∈ j1, . . . , jk we have

∣∣(∂ j1 · · ·∂ jk m)(ξ1, . . . ,ξn)∣∣≤ A |ξ j1 |−1 · · · |ξ jk |−1 . (6.2.9)

Then m satisfies (6.2.6).

Proof. Simply observe that condition (6.2.9) implies (6.2.5).


Example 6.2.6. The following are examples of functions that satisfy the hypotheses

of Corollary 6.2.5:

m1(ξ ) =ξ1

ξ1 + i(ξ 22 + · · ·+ξ 2

n ),

m2(ξ ) =|ξ1|α1 · · · |ξn|αn

(ξ 21 +ξ 2

2 + · · ·+ξ 2n )

α/2,

where α1 +α2 + · · ·+αn = α , α j > 0,

m3(ξ ) =ξ2ξ

23

iξ1 +ξ 22 +ξ 4

3

.

The functions m1 and m2 are defined on Rn \0 and m3 on R3 \0.The previous examples and many other examples that satisfy the hypothesis

(6.2.9) of Corollary 6.2.5 are invariant under a set of dilations in the following sense:

suppose that there exist k1, . . . ,kn ∈ R+ and s ∈ R such that the smooth function m

on Rn \0 satisfies

m(λ k1ξ1, . . . ,λknξn) = λ ism(ξ1, . . . ,ξn)

for all ξ1, . . . ,ξn ∈ R and λ > 0. Then m satisfies condition (6.2.9). Indeed, differ-

entiation gives

λα1k1+···+αnkn∂αm(λ k1ξ1, . . . ,λknξn) = λ is∂αm(ξ1, . . . ,ξn)

for every multi-index α = (α1, . . . ,αn). Now for every ξ ∈Rn \0 pick the unique

λξ > 0 such that (λ k1

ξξ1, . . . ,λ

kn

ξξn) ∈ Sn−1. Then λ

k jα j

ξ≤ |ξ j|−α j , and it follows

that

|∂αm(ξ1, . . . ,ξn)| ≤[

supSn−1

|∂αm|]λα1k1+···+αnkn

ξ≤Cα |ξ1|−α1 · · · |ξn|−αn .

6.2.3 The Mihlin–Hormander Multiplier Theorem on Rn

We now discuss another multiplier theorem that also requires decay of derivatives.

We will consider the situation where each differentiation produces uniform decay in

all variables, quantitatively expressed via the condition

|∂αξ m(ξ )| ≤Cα |ξ |−|α | (6.2.10)

for each multi-index α . The decay can also be expressed in terms of a square inte-

grable estimate that has the form

(∫

R<|ξ |<2R|∂αξ m(ξ )|2 dξ

) 12

≤C′α Rn2−|α | < ∞ (6.2.11)

for all multi-indices α and all R > 0. Obviously (6.2.10) implies (6.2.11)

Theorem 6.2.7. Let m(ξ ) be a complex-valued bounded function on Rn \ 0 that

satisfies for some A < ∞

(∫

R<|ξ |<2R|∂αξ m(ξ )|2 dξ

) 12

≤ ARn2−|α | < ∞ (6.2.12)

for all multi-indices |α| ≤ [n/2]+1 and all R > 0.

Then for all 1 < p < ∞, m lies in Mp(Rn) and the following estimate is valid:

‖m‖Mp≤Cn max(p,(p−1)−1)

(A+‖m‖L∞

). (6.2.13)

Moreover, the operator f → ( f m)∨ maps L1(Rn) to L1,∞(Rn) with norm at most a

dimensional constant multiple of A+‖m‖L∞ .

We remark that in most applications, condition (6.2.12) appears in the form

|∂αξ m(ξ )| ≤Cα |ξ |−|α | , (6.2.14)

which should be, in principle, easier to verify.

Proof. Since m is a bounded function, the operator given by convolution with W =m∨ is bounded on L2(Rn). To prove that this operator maps L1(Rn) to L1,∞(Rn), it

suffices to prove that the distribution W coincides with a function K on Rn \0 that

satisfies Hormander’s condition.

Let ζ be a smooth function supported in the annulus 12≤ |ξ | ≤ 2 such that

∑j∈Z

ζ (2− jξ ) = 1, when ξ = 0.

Set m j(ξ ) = m(ξ )ζ (2− jξ ) for j ∈ Z and K j = m∨j . We begin by observing that

∑N−N K j converges to W in S ′(Rn). Indeed, for all ϕ ∈S (Rn) we have

⟨ N

∑j=−N

K j,ϕ⟩=

⟨ N

∑j=−N

m j,ϕ∨⟩→

⟨m,ϕ ∨

⟩=

⟨W,ϕ

⟩.

We set n0 = [ n2]+1. We claim that there is a constant Cn such that

supj∈Z

∫

Rn|K j(x)|(1+2 j|x|) 1

4 dx ≤ CnA , (6.2.15)

supj∈Z

2− j

∫

Rn|∇K j(x)|(1+2 j|x|) 1

4 dx ≤ CnA . (6.2.16)

To prove (6.2.15) we multiply and divide the integrand in (6.2.15) by the expression

(1+ 2 j|x|)n0 . Applying the Cauchy–Schwarz inequality to |K j(x)|(1+ 2 j|x|)n0 and

(1+2 j|x|)−n0+14 , we control the integral in (6.2.15) by the product

(∫

Rn|K j(x)|2(1+2 j|x|)2n0 dx

)12(∫

Rn(1+2 j|x|)−2n0+

12 dx

)12

. (6.2.17)

We now note that −2n0 +12< −n, and hence the second factor in (6.2.17) is equal

to a constant multiple of 2− jn/2. To estimate the first integral in (6.2.17) we use the

simple fact that

(1+2 j|x|)n0 ≤C(n) ∑|γ |≤n0

|(2 jx)γ | .

We now have that the expression inside the supremum in (6.2.15) is controlled by

C′(n)2− jn/2 ∑|γ |≤n0

(∫

Rn|K j(x)|222 j|γ ||xγ |2 dx

)12

, (6.2.18)

which, by Plancherel’s theorem, is equal to

2− jn/2 ∑|γ |≤n0

Cγ2j|γ |

(∫

Rn|(∂ γm j)(ξ )|2 dξ

)12

(6.2.19)

for some constants Cγ .

For multi-indices δ = (δ1, . . . ,δn) and γ = (γ1, . . . ,γn) we introduce the notation

δ ≤ γ to mean δ j ≤ γ j for all j = 1, . . . ,n . For any |γ | ≤ n0 we use Leibniz’s rule to

obtain for some constants Cδ ,γ

(∫

Rn|(∂ γm j)(ξ )|2 dξ

)12

≤ ∑δ≤γ

Cδ ,γ

(∫

Rn

∣∣2− j|γ−δ |(∂ γ−δξ

ζ )(2− jξ )(∂ δξ m)(ξ )∣∣2dξ

)12

≤ ∑δ≤γ

Cδ ,γ2− j|γ |2 j|δ |

(∫

2 j−1≤|ξ |≤2 j+1|(∂ δξ m)(ξ )

∣∣2 dξ

)12

≤ ∑δ≤γ

Cδ ,γ2− j|γ |2 j|δ | 2A2 jn/22− j|δ |

=Cn A2 jn/22− j|γ | ,

which inserted in (6.2.19) and combined with (6.2.18) yields (6.2.15). To obtain

(6.2.16) we repeat the same argument for every derivative ∂rK j. Since the Fourier

transform of (∂rK j)(x)xγ is equal to a constant multiple of ∂ γ(ξrm(ξ )ζ (2− jξ )

),

we observe that the extra factor 2− j in (6.2.16) can be combined with ξr to write

2− j∂ γ(ξrm(ξ )ζ (2− jξ )

)as ∂ γ

(m(ξ )ζr(2

− jξ )), where ζr(ξ ) = ξrζ (ξ ). The pre-

vious calculation with ζr replacing ζ can then be used to complete the proof of

(6.2.16).

We now show that for all x = 0, the series ∑ j∈Z K j(x) converges to a function,

which we denote by K(x). Indeed, as a consequence of (6.2.15) we have that

(1+2 jδ )14

∫

|x|≥δ|K j(x)|dx≤ CnA ,

for any δ > 0, which implies that the function ∑ j>0 |K j(x)| is integrable away from

the origin and satisfies∫δ≤|x|≤2δ ∑ j>0 |K j(x)|dx < ∞. Now note that (6.2.15) also

holds with − 14

in place of 14. Using this observation we obtain

(1+2 j2δ )−14

∫

|x|≤2δ|K j(x)|dx≤

∫

|x|≤2δ|K j(x)|(1+2 j|x|)− 1

4 dx≤ CnA ,

and from this it follows that∫δ≤|x|≤2δ ∑ j≤0 |K j(x)|dx < ∞.

We conclude that the series ∑ j∈Z K j(x) converges a.e. on Rn \0 to a function

K(x) that coincides with the distribution W = m∨ on Rn \0 and satisfies

supδ>0

∫

δ≤|x|≤2δ|K(x)|dx < ∞ .

We now prove that the function K = ∑ j∈Z K j (defined on Rn \ 0) satisfies

Hormander’s condition. It suffices to prove that for all y = 0 we have

∑j∈Z

∫

|x|≥2|y||K j(x− y)−K j(x)|dx≤ 2C′nA . (6.2.20)

Fix a y ∈Rn \0 and pick a k ∈ Z such that 2−k ≤ |y| ≤ 2−k+1. The part of the sum

in (6.2.20) where j > k is bounded by

∑j>k

∫

|x|≥2|y||K j(x− y)|+ |K j(x)|dx ≤ 2∑

j>k

∫

|x|≥|y||K j(x)|dx

≤ 2∑j>k

∫

|x|≥|y||K j(x)|

(1+2 j|x|) 14

(1+2 j|x|) 14

dx

≤ ∑j>k

2CnA

(1+2 j|y|) 14

≤ ∑j>k

2CnA

(1+2 j2−k)14

=C′nA ,

where we used (6.2.15). The part of the sum in (6.2.20) where j ≤ k is bounded by

∑j≤k

∫

|x|≥2|y||K j(x− y)−K j(x)|dx

≤ ∑j≤k

∫

|x|≥2|y|

∫ 1

0|− y ·∇K j(x−θy)|dθ dx

≤∫ 1

0∑j≤k

2−k+1∫

Rn|∇K j(x−θy)|(1+2 j|x−θy|) 1

4 dxdθ

≤∫ 1

0∑j≤k

2−k+1CnA2 j dθ ≤C′nA ,

using (6.2.16). Hormander’s condition is satisfied for K, and we appeal to Theorem

5.3.3 to complete the proof of (6.2.13).

Example 6.2.8. Let m be a smooth function away from the origin that is homoge-

neous of imaginary order, i.e., for some fixed τ real and all λ > 0 we have

m(λξ ) = λ iτm(ξ ) . (6.2.21)

Then m is an Lp Fourier multiplier for 1 < p < ∞. Indeed, differentiating both sides

of (6.2.21) with respect to ∂αξ

we obtain

λ |α |∂αξ m(λξ ) = λ iτ∂αξ m(ξ )

and taking λ = |ξ |−1, we deduce condition (6.2.14) with Cα = sup|θ |=1 |∂αm(θ)|.An explicit example of such a function is m(ξ ) = |ξ |iτ . Another example is

m0(ξ1,ξ2,ξ3) =ξ 2

1 +ξ 22

ξ 21 + i(ξ 2

2 +ξ 23 )

which is homogeneous of degree zero and also smooth on Rn \0.Example 6.2.9. Let z be a complex numbers with Rez≥ 0. Then the functions

m1(ξ ) =

( |ξ |21+ |ξ |2

)z

, m2(ξ ) =

(1

1+ |ξ |2)z

defined on Rn are Lp Fourier multipliers for 1 < p < ∞. To prove this assertion for

m1, we verify condition (6.2.14). To achieve this, introduce the function on Rn+1

M1(ξ1, . . . ,ξn, t) =

( |ξ1|2 + · · ·+ |ξn|2t2 + |ξ1|2 + · · ·+ |ξn|2

)z

=

( |ξ |2t2 + |ξ |2

)z

,

where ξ = (ξ1, . . . ,ξn). Then M is homogeneous of degree 0 and smooth on

Rn+1 \0. The derivatives ∂ βM1 are homogeneous of degree −|β | and by the cal-

culation in the preceding example they satisfy |∂ βM1(ξ , t)| ≤ Cβ |(ξ , t)|−|β |, with

Cβ = sup|θ |=1 |∂ βM1(θ)|, whenever (ξ , t) = 0 and β is a multi index of n+1 vari-

ables. In particular, taking β = (α,0), we obtain

∣∣∂α1

ξ1· · ·∂αn

ξnM1(ξ1, . . . ,ξn, t)

∣∣≤ Cα

(t2 + |ξ |2)|α |/2,

and setting t = 1 we deduce that |∂αm1(ξ )| ≤Cα(1+ |ξ |2)−|α |/2 ≤Cα |ξ |−|α |.

For m2 we introduce the function

M2(ξ1, . . . ,ξn, t) =

(1

t2 + |ξ1|2 + · · ·+ |ξn|2)z

on Rn+1, which is homogeneous of degree−2z. Then the derivative ∂ βM2 is homo-

geneous of degree −|β | − 2z, hence it satisfies |∂ βM2(ξ , t)| ≤ Cβ |(ξ , t)|−|β |−2Rez

for all multi-indices β of n+1 variables. In particular, taking β = (α,0), we obtain

∣∣∂α1

ξ1· · ·∂αn

ξnM2(ξ1, . . . ,ξn, t)

∣∣≤ Cα

(t2 + |ξ |2) |α|2 +Rez,

and setting t = 1, we deduce |∂αm2(ξ )| ≤Cα(1+ |ξ |2)−|α |/2 ≤Cα |ξ |−|α |, where in

the first inequality we used that Rez≥ 0.

We end this section by comparing Theorems 6.2.2 and 6.2.4 with Theorem 6.2.7.

It is obvious that in dimension n = 1, Theorem 6.2.2 is stronger than Theorem 6.2.7

in view of the inequality

∫

2 j<|ξ |<2 j+1|m′(ξ )|dξ ≤ 2 j/2

(∫

2 j<|ξ |<2 j+1|m′(ξ )|2 dξ

) 12

,

which implies that (6.2.2) is weaker than (6.2.12). Note also that in Theorem 6.2.2

the multiplier m is not required to be differentiable at the points ±2 j. But in higher

dimensions neither theorem includes the other. In Theorem 6.2.4 the multiplier is

allowed to be singular on a set of measure zero but is required to be differentiable in

every variable, i.e., to be at least C n in the complement of this null set. In Theorem

6.2.7, the multiplier is only allowed to be singular only at the origin, but it is assumed

to be C [n/2]+1, requiring almost half the differentiability called for by condition

(6.2.9). It should be noted that both theorems have their shortcomings. In particular,

they are not Lp sensitive, i.e., delicate enough to detect whether m is a bounded

Fourier multiplier on some Lp but not on some other Lq.

Exercises

6.2.1. Let ψ(ξ ) be a smooth function supported in [3/4,2]∪ [−2,−3/4] and equal

to 1 on [1,3/2]∪ [−3/2,−1] that satisfies ∑ j∈Zψ(2− jξ ) = 1 for all ξ = 0. Let

1 ≤ k ≤ n. Prove that m ∈Mp(Rn) if and only if (6.2.1) is satisfied with mj(ξ ) re-

placed by the function m(ξ )ψ(2− j1ξ1) · · ·ψ(2− jkξk).[Hint: To prove one direction, partition Zk in 2k sets such that for every j =( j1, . . . , jk) in each of these sets, ji has a fixed remainder modulo 2. For the

other direction, use Theorem 6.1.6 in the variables x1, . . . ,xk. Also use the inequal-

ity ‖ f‖Lp(Rn) ≤Cp‖(∑j∈Zk |( f χRj)∨|2)1/2‖Lp(Rn), Rj = ([−2,− 1

2]∪ [ 1

2,2])k×Rn−k,

which can be derived by duality from the identity ∑j∈Zk χRj= 2k.

]

6.2.2. Let ϕ be a smooth function on the real line supported in the interval [−1,1].Let ψ(t) be a smooth function on the real line that is equal to 1 when |t| ≥ 10 and

vanishes when |t| ≤ 9. Show that for the function m(ξ1,ξ2) = eiξ 22 /ξ1ϕ(ξ2)ψ(ξ1)

lies in Mp(R2), 1 < p < ∞, using Theorem 6.2.4. Also show that Theorem 6.2.7

does not apply.

6.2.3. Consider the differential operators

L1 = ∂1−∂ 22 +∂ 4

3 ,

L2 = ∂1 +∂ 22 +∂ 2

3 .

Prove that for every 1 < p < ∞ there exists a constant Cp < ∞ such that for all

Schwartz functions f on R3 we have

∥∥∂2∂23 f

∥∥Lp ≤Cp

∥∥L1( f )∥∥

Lp ,∥∥∂1 f∥∥

Lp ≤Cp

∥∥L2( f )∥∥

Lp .

[Hint: Use Corollary 6.2.5 and the idea of Example 6.2.6.

]

6.2.4. Suppose that m(ξ ) is a real-valued function that satisfies either (6.2.9) or

|∂αm(ξ )| ≤Cα |ξ |−|α | for all multi-indices α with |α| ≤ [ n2]+1 and all ξ ∈Rn\0.

Show that eim(ξ ) lies in Mp(Rn) for any 1 < p < ∞.[

Hint: Prove by induction and use that

∂α(eim(ξ )

)= eim(ξ ) ∑

l j≥0,β j≤αl1β

1+···+lkβk=α

cβ 1,...,β k(∂ β1

m(ξ ))l1 · · ·(∂ β k

m(ξ ))lk ,

where the sum is taken over all partitions of the multi-index α as a linear combina-

tion of multi-indices β j with coefficients l j ∈ Z+∪0.]

6.2.5. Suppose that ϕ(ξ ) is a smooth function on Rn that vanishes in a neighbor-

hood of the origin and is equal to 1 in a neighborhood of infinity. Prove that the

function eiξ j |ξ |−1ϕ(ξ ) is in Mp(R

n) for 1 < p < ∞.

6.2.6. Let τ ,τ1, . . . ,τn be real numbers and ρ1, . . . ,ρn be even natural numbers.

Prove that the following functions are Lp multipliers on Rn for 1 < p < ∞:

|ξ1|iτ1 · · · |ξn|iτn ,

(|ξ1|ρ1 + · · ·+ |ξn|ρn)iτ ,

(|ξ1|−ρ1 + |ξ2|−ρ2)iτ .

6.2.7. Let ζ (ξ ) be a smooth function on Rn is supported in a compact set that does

not contain the origin and let a j be a bounded sequence of complex numbers. Prove

that the function

m(ξ ) = ∑j∈Z

a jζ (2− jξ )

is in Mp(Rn) for all 1 < p < ∞.

6.2.8. Let ζ (ξ ) be a smooth function on Rn supported in a compact set that does

not contain the origin and let ∆ζj ( f ) =

(f (ξ )ζ (2− jξ )

)∨. Show that the operator

f → supN∈Z

∣∣∣ ∑j<N

∆ζj ( f )

∣∣∣

is bounded on Lp(R) when 1 < p < ∞.[Hint: Pick a Schwartz function ϕ satisfying ∑ j∈Z ϕ(2

− jξ ) = 1 on Rn \ 0 with

ϕ(ξ ) supported in 67≤ |ξ | ≤ 2. Then ∆

ϕk ∆

ζj = 0 if | j− k|< c0 and we have

∑j<N

∆ζj = ∑

k<N+c0

∆ϕk ∑

j<N

∆ζj = ∑

k<N+c0

∆ϕk ∑

j

∆ζj − ∑

k<N+c0

∆ϕk ∑

j≥N

∆ζj ,

which is a finite sum plus a term controlled by a multiple of the operator

f →M(∑j∈Z

∆ζj ( f )

),

where M is the Hardy–Littlewood maximal function.]

6.2.9. Let Ψ be a Schwartz function whose Fourier transform is real-valued, sup-

ported in a compact set that does not contain the origin, and satisfies

∑j∈Z

Ψ(2− jξ ) = 1 when ξ = 0.

Let ∆ j be the Littlewood–Paley operator associated withΨ . Prove that

∥∥ ∑| j|<N

∆ j(g)−g∥∥

Lp → 0

as N → ∞ for all functions g ∈ S (Rn). Deduce that Schwartz functions whose

Fourier transforms have compact supports that do not contain the origin are dense

in Lp(Rn) for 1 < p < ∞.[Hint: Use the result of Exercise 6.2.8 and the Lebesgue dominated convergence

theorem.]

6.3 Applications of Littlewood–Paley Theory 453

6.3 Applications of Littlewood–Paley Theory

We now turn our attention to some important applications of Littlewood–Paley the-

ory. We are interested in obtaining bounds for singular and maximal operators.

These bounds are obtained by controlling the corresponding operators by quadratic

expressions.

6.3.1 Estimates for Maximal Operators

One way to control the maximal operator supk |Tk( f )| is by introducing a good aver-

aging function ϕ and using the majorization

supk

|Tk( f )| ≤ supk

|Tk( f )− f ∗ϕ2−k |+ supk

| f ∗ϕ2−k |

≤(∑k

|Tk( f )− f ∗ϕ2−k |2) 1

2+Cϕ M( f )

(6.3.1)

for some constant Cϕ depending on ϕ . We apply this idea to prove the following

theorem.

Theorem 6.3.1. Let m be a bounded function on Rn that is C 1 in a neighborhood

of the origin and satisfies m(0) = 1 and |m(ξ )| ≤ C|ξ |−ε for some C,ε > 0 and

all ξ = 0. For each k ∈ Z define Tk( f )(x) = ( f (ξ )m(2−kξ ))∨(x). Then there is a

constant Cn such that for all L2 functions f on Rn we have

∥∥supk∈Z

|Tk( f )|∥∥

L2 ≤Cn

∥∥ f∥∥

L2 . (6.3.2)

Proof. Select a Schwartz function ϕ such that ϕ(0) = 1. Then there are positive

constants C1 and C2 such that |m(ξ )− ϕ(ξ )| ≤C1|ξ |−ε for |ξ | away from zero and

|m(ξ )− ϕ(ξ )| ≤C2|ξ | for |ξ | near zero. These two inequalities imply that

∑k

|m(2−kξ )− ϕ(2−kξ )|2 ≤C3 < ∞ ,

from which the L2 boundedness of the operator

f →(∑k

|Tk( f )− f ∗ϕ2−k |2)1/2

follows easily. Using estimate (6.3.1) and the well-known L2 estimate for the Hardy–

Littlewood maximal function, we obtain (6.3.2).

If m(ξ ) is the characteristic function of a rectangle with sides parallel to the axes,

this result can be extended to Lp.

Theorem 6.3.2. Let 1 < p <∞ and let U be the n-fold product of open intervals that

contain zero. For each k ∈ Z define Tk( f )(x) = ( f (ξ )χU(2−kξ ))∨(x). Then there is

a constant Cp,n such that for all Lp functions f on Rn we have

∥∥supk∈Z

|Tk( f )|∥∥

Lp(Rn)≤Cp,n

∥∥ f∥∥

Lp(Rn).

Proof. Let us fix an open annulus A whose interior contains the boundary of U and

take a smooth function with compact support ψ that vanishes in a neighborhood of

zero and a neighborhood of infinity and is equal to 1 on the annulus A. Then the

function ϕ = (1− ψ )χU is Schwartz. Since χU = χU ψ + ϕ , it follows that for all

f ∈ Lp(Rn) we have

Tk( f ) = Tk( f )− f ∗ϕ2−k + f ∗ϕ2−k = Tk( f ∗ψ2−k)+ f ∗ϕ2−k .

Taking the supremum over k and using Corollary 2.1.12 we obtain

supk∈Z

|Tk( f )| ≤(∑k

|Tk( f )− f ∗ϕ2−k |2)1/2

+Cϕ M( f ) . (6.3.3)

The operator Tk( f )− f ∗ϕ2−k is given by multiplication on the Fourier transform

side by the multiplier

χU (2−kξ )− ϕ(2−kξ ) = χU (2

−kξ )ψ(2−kξ ) = χ2kU (ξ )ψ(2−kξ ) .

Since 2kUk∈Z is a measurable family of rectangles with sides parallel to the axes,

Exercise 5.6.1(b) yields the following inequality:

∥∥∥(∑k∈Z

|Tk( f )− f ∗ϕ2−k |2) 1

2

∥∥∥Lp≤Cp,n

∥∥∥(∑k∈Z

| f ∗ψ2−k |2) 1

2

∥∥∥Lp. (6.3.4)

Since f ∗ψ2−k = ∆ψj ( f ), estimate (6.1.4) of Theorem 6.1.2 yields that the expres-

sion on the right in (6.3.4) is controlled by a multiple of ‖ f‖Lp . Taking Lp norms in

(6.3.3) and using the Lp estimate for the square function yields the required conclu-

sion.

The following lacunary version of the Carleson–Hunt theorem is yet another in-

dication of the powerful techniques of Littlewood–Paley theory.

Corollary 6.3.3. (a) Let f be in L2(Rn) and let Ω be an open set that contains the

origin in Rn. Then

limk→∞

∫

2kΩf (ξ )e2πix·ξ dξ = f (x)

for almost all x ∈ Rn.

(b) Let f be in Lp(Rn) for some 1 < p < ∞. Then

limk→∞

∫|ξ1|<2k

...|ξn|<2k

f (ξ )e2πix·ξ dξ = f (x)

Proof. Both limits exist everywhere for functions f in the Schwartz class. To obtain

almost everywhere convergence for general f in Lp we appeal to Theorem 2.1.14.

The required control of the corresponding maximal operator is a consequence of

Theorem 6.3.1 with m = χΩ in case (a) and Theorem 6.3.2 in case (b).

6.3.2 Estimates for Singular Integrals with Rough Kernels

We now turn to another application of the Littlewood–Paley theory involving singu-

lar integrals.

Theorem 6.3.4. Suppose that µ is a finite Borel measure on Rn with compact sup-

port that satisfies |µ(ξ )| ≤ Bmin(|ξ |−b, |ξ |b

)for some b > 0 and all ξ = 0. Define

measures µ j by setting µ j(ξ ) = µ(2− jξ ). Then the operator

Tµ( f )(x) = ∑j∈Z

( f ∗µ j)(x)

is bounded on Lp(Rn) for all 1 < p < ∞.

Proof. It is natural to begin with the L2 boundedness of Tµ . The estimate on µimplies that

∑j∈Z

|µ(2− jξ )| ≤ ∑j∈Z

Bmin(|2− jξ |b, |2− jξ |−b

)≤CbB < ∞ . (6.3.5)

The L2 boundedness of Tµ is an immediate consequence of (6.3.5).

We now turn to the Lp boundedness of Tµ for 1< p<∞. We fix a radial Schwartz

function ψ whose Fourier transform is supported in the annulus 12< |ξ | < 2 that

satisfies

∑j∈Z

ψ(2− jξ ) = 1 (6.3.6)

whenever ξ = 0. We let ψ2−k(x) = 2knψ(2kx), so that ψ2−k(ξ ) = ψ(2−kξ ), and we

observe that the identity

µ j = ∑k∈Z

µ j ∗ψ2− j−k

is valid by taking Fourier transforms and using (6.3.6). We now define operators Sk

by setting

Sk( f ) = ∑j∈Z

µ j ∗ψ2− j−k ∗ f = ∑j∈Z

(µ ∗ψ2−k)2− j ∗ f .

Then for f in S we have that

Tµ( f ) = ∑j∈Z

µ j ∗ f = ∑j∈Z∑k∈Z

µ j ∗ψ2− j−k ∗ f = ∑k∈Z

Sk( f ) .


It suffices therefore to obtain Lp boundedness for the sum of the Sk’s. We begin

by investigating the L2 boundedness of each Sk. Since the product ψ2− j−k ψ2− j′−k is

nonzero only when j′ ∈ j−1, j, j+1, it follows that

∥∥Sk( f )∥∥2

L2 ≤ ∑j∈Z∑j′∈Z

∫

Rn|µ j(ξ )µ j′ (ξ )ψ(2

− j−kξ )ψ(2− j′−kξ )| | f (ξ )|2 dξ

≤C1 ∑j∈Z

j+1

∑j′= j−1

∫

|ξ |≈2 j+k

|µ j(ξ )µ j′(ξ )| | f (ξ )|2 dξ

≤C2 ∑j∈Z

∫

|ξ |≈2 j+k

B2 min(|2− jξ |b, |2− jξ |−b)2| f (ξ )|2 dξ

≤C23B22−2|k|b ∑

j∈Z

∫

|ξ |≈2 j+k

| f (ξ )|2 dξ

=C23B2 2−2|k|b∥∥ f

∥∥2

L2 .

We have therefore obtained that for all k ∈ Z and f ∈S (Rn) we have

∥∥Sk( f )∥∥

L2 ≤C3 B2−b|k|∥∥ f∥∥

L2 . (6.3.7)

We notice that for any R > 0 we have

∫

R≤|x|≤2R∑j∈Z

∣∣(µ ∗ψ2−k

)2− j(x)

∣∣dx = ∑j∈Z

∫

2 jR≤|x|≤2 j+1R

∣∣(µ ∗ψ2−k

)(x)

∣∣dx

=∫

Rn

∣∣(µ ∗ψ2−k

)(x)

∣∣dx

≤∥∥µ

∥∥‖ψ‖L1 ,

thus condition (5.3.4) of Theorem 5.3.3 is satisfied.

Next we verify that the kernel of each Sk satisfies Hormander’s condition with

constant at most a multiple of (1+ |k|). Fix y = 0. Then

∫

|x|≥2|y|

∣∣∣∣∑j∈Z

((µ ∗ψ2−k)2− j(x− y)− (µ ∗ψ2−k)2− j(x)

)∣∣∣∣dx

≤ ∑j∈Z

∫

|x|≥2|y|2 jn

∣∣(µ ∗ψ2−k)(2 jx−2 jy)− (µ ∗ψ2−k)(2 jx)∣∣dx

= ∑j∈Z

I j,k(y) ,

where

I j,k(y) =∫

|x|≥2 j+1|y|

∣∣(µ ∗ψ2−k)(x−2 jy)− (µ ∗ψ2−k)(x)∣∣dx .

We observe that I j,k(y) ≤C4‖µ‖M . Let |µ | be the total variation of µ . To obtain a

more delicate estimate for I j,k(y) we argue as follows:

I j,k(y) ≤∫

|x|≥2 j+1|y|

∫

Rn

∣∣ψ2−k(x−2 jy− z)−ψ2−k(x− z)∣∣d|µ |(z)dx

=∫

Rn2kn

∫

|x|≥2 j+1|y|

∣∣ψ(2kx−2kz−2 j+ky)−ψ(2kx−2kz)∣∣dxd|µ |(z)

≤C5

∫

|x|≥2 j+1|y|

∫

Rn2kn2 j+k|y|

∣∣∇ψ(2kx−2kz−θ)∣∣d|µ |(z)dx

≤C62 j+k

∫

Rn

∫

|x|≥2 j+1|y|

2kn|y|(1+ |2kx−2kz−θ |

)−n−2dxd|µ |(z)

=C62 j+k|y|∫

Rn

∫

|x|≥2 j+k+1|y|

(1+ |x−2kz−θ |

)−n−2dxd|µ |(z) ,

where |θ | ≤ 2 j+k|y|. Note that θ depends on j,k, and y. From this and from I j,k(y)≤C4‖µ‖M we obtain

I j,k(y)≤C7

∥∥µ∥∥

Mmin

(1,2 j+k|y|

), (6.3.8)

which is valid for all j,k, and y = 0. To estimate the last double integral even more

delicately, we consider the following two cases: |x| ≥ 2k+2|z| and |x| < 2k+2|z|. In

the first case we have |x− 2kz− θ | ≥ 14|x|, given the fact that |x| ≥ 2 j+k+1|y|. In

the second case we have that |x| ≤ 2k+2R, where B(0,R) contains the support of

µ . Applying these observations in the last double integral, we obtain the following

estimate:

I j,k(y)≤ C82 j+k|y|∫

Rn

[ ∫

|x|≥2 j+k+1|y||x|≥2k+2|z|

dx(1+ 1

4|x|

)n+2+

∫

|x|≥2 j+k+1|y||x|<2k+2R

dx

]d|µ |(z)

≤ C92 j+k|y|∥∥µ

∥∥M

[1

(2 j+k|y|)2+0

]

= C9(2j+k|y|)−1

∥∥µ∥∥

M,

provided 2 j|y| ≥ 2R. Combining this estimate with (6.3.8), we obtain

I j,k(y)≤C10

∥∥µ∥∥

M

min

(1,2 j+k|y|

)for all j,k and y,

(2 j+k|y|)−1 when 2 j|y| ≥ 2R.(6.3.9)

We now estimate ∑ j I j,k(y). When 2k ≥ (2R)−1 we use (6.3.9) to obtain

∑j

I j,k(y) ≤ C10

∥∥µ∥∥

M

[∑

2 j≤ 1

2k |y|

2 j+k|y|+ ∑1

2k |y|≤2 j≤ 2R|y|

1+ ∑2 j≥ 2R

|y|

(2 j+k|y|)−1]

≤C11

∥∥µ∥∥

M(| logR|+ |k|) .

Also when 2k < (2R)−1 we again use (6.3.9) to obtain

∑j

I j,k(y)≤C10

∥∥µ∥∥

M

[∑

2 j≤ 1

2k |y|

2 j+k|y|+ ∑2 j≥ 1

2k |y|

(2 j+k|y|)−1]≤C12

∥∥µ∥∥

M,

since in the second sum we have 2 j|y| ≥ 2−k > 2R, which justifies use of the corre-

sponding estimate in (6.3.9). This gives

∑j

I j,k(y)≤C13

∥∥µ∥∥

M(1+ |k|) , (6.3.10)

where the constant C13 depends on the dimension and on R. We now use esti-

mates (6.3.7) and (6.3.10) and Theorem 5.3.3 to obtain that each Sk maps L1(Rn)to L1,∞(Rn) with constant at most

Cn(2−b|k|+1+ |k|)

∥∥µ∥∥

M≤Cn(2+ |k|)

∥∥µ∥∥

M.

It follows from the Marcinkiewicz interpolation theorem (Theorem 1.3.2) that Sk

maps Lp(Rn) to itself for 1 2 follows by duality.

An immediate consequence of the previous result is the following.

Corollary 6.3.5. Suppose that µ is a finite Borel measure on Rn with compact sup-

port that satisfies |µ(ξ )| ≤ Bmin(|ξ |−b, |ξ |b

)for some b > 0 and all ξ = 0. Define

measures µ j by setting µ j(ξ ) = µ(2− jξ ). Then the square function

G( f ) =(∑j∈Z

|µ j ∗ f |2) 1

2(6.3.11)

maps Lp(Rn) to itself whenever 1 < p < ∞.

Proof. To obtain the boundedness of the square function in (6.3.11) we use the

Rademacher functions r j(t), introduced in Appendix C.1, reindexed so that their

index set is the set of all integers (not the set of nonnegative integers). For each t we

introduce the operators

T tµ( f ) = ∑

j∈Z

r j(t)( f ∗µ j) .

Next we observe that for each t in [0,1] the operators T tµ map Lp(Rn) to itself with

the same constant as the operator Tµ , which is in particular independent of t. Using

that the square function in (6.3.11) raised to the power p is controlled by a multiple

of the quantity ∫ 1

0

∣∣∣∑j∈Z

r j(t)( f ∗µ j)∣∣∣

p

dt ,

a fact stated in Appendix C.2, we obtain the required conclusion by integrating

over Rn.

6.3.3 An Almost Orthogonality Principle on Lp

Suppose that Tj are multiplier operators given by Tj( f ) = ( f m j)∨, for some multi-

pliers m j. If the functions m j have disjoint supports and they are bounded uniformly

in j, then the operator

T =∑j

Tj

is bounded on L2. The following theorem gives an Lp analogue of this result.

Theorem 6.3.6. Suppose that 1 < p ≤ 2 ≤ q < ∞. Let m j be Schwartz functions

supported in the annuli 2 j−1 ≤ |ξ | ≤ 2 j+1 and let Tj( f ) = ( f m j)∨. Suppose that the

Tj’s are uniformly bounded operators from Lp(Rn) to Lq(Rn), i.e.,

supj

∥∥Tj

∥∥Lp→Lq = A < ∞ .

Then for each f ∈ Lp(Rn), the series

T ( f ) =∑j

Tj( f )

converges in the Lq norm and there exists a constant Cp,q,n < ∞ such that

∥∥T∥∥

Lp→Lq ≤Cp,q,nA. (6.3.12)

Proof. Fix a radial Schwartz function ϕ whose Fourier transform ϕ is real, equal to

one on the annulus 12≤ |ξ | ≤ 2, and vanishes outside the annulus 1

4≤ |ξ | ≤ 4. We

set ϕ2− j(x) = 2 jnϕ(2 jx), so that ϕ2− j is equal to 1 on the support of each m j. Setting

∆ j( f ) = f ∗ϕ2− j , we observe that

Tj = ∆ jTj∆ j

for all j ∈ Z. For a positive integer N we set

T N = ∑| j|≤N

∆ jTj∆ j .


Fix f ∈ Lp(Rn). Clearly for every N, T N( f ) is in Lq(Rn). Using (6.1.21) we obtain

∥∥T N( f )∥∥

Lq =∥∥ ∑| j|≤N

∆ jTj∆ j( f )∥∥

Lq

≤C′q

∥∥∥(∑j∈Z

|Tj∆ j( f )|2) 1

2

∥∥∥Lq

=C′q

∥∥∥∑j∈Z

|Tj∆ j( f )|2∥∥∥

12

Lq/2

≤C′q(∑j∈Z

∥∥∥|Tj∆ j( f )|2∥∥∥

Lq/2

) 12

=C′q(∑j∈Z

∥∥Tj∆ j( f )∥∥2

Lq

) 12,

where we used Minkowski’s inequality, since q/2≥ 1. Using the uniform bounded-

ness of the Tj’s from Lp to Lq, we deduce that

C′q(∑j∈Z

∥∥Tj∆ j( f )∥∥2

Lq

) 12 ≤C′q A

(∑j∈Z

∥∥∆ j( f )∥∥2

Lp

)12

=C′q A(∑j∈Z

∥∥|∆ j( f )|2∥∥

Lp/2

)12

≤C′q A(∥∥∥∑

j∈Z

|∆ j( f )|2∥∥∥

Lp/2

)12

=C′q A

∥∥∥(∑j∈Z

|∆ j( f )|2) 1

2∥∥∥

Lp

≤C′q Cp A∥∥ f

∥∥Lp(Rn)

,

where we used the result of Exercise 1.1.5(b), since p ≤ 2, and Theorem 6.1.2. We

conclude that the operators T N are uniformly bounded from Lp(Rn) to Lq(Rn).

If h is compactly supported in a subset of Rn \ 0, then the sequence T N(h)becomes independent of N for N large enough and hence it is Cauchy in Lq. But in

view of Exercise 6.2.9, the set of all such h is dense in Lp(Rn). Combining these

two results with the uniform boundedness of the T N’s from Lp to Lq, a simple ε3

argument gives that for all f ∈ Lp the sequence T N( f ) is Cauchy in Lq. Therefore,

for all f ∈ Lp the sequence T N( f )N converges in Lq to some T ( f ). Fatou’s lemma

gives ∥∥T ( f )∥∥

Lq ≤C′q Cp A∥∥ f

∥∥Lp ,

which proves (6.3.12).

Exercises

6.3.1. (The g-function) Let Pt(x) = Γ ( n+12)π−

n+12 t(t2 + |x|2)− n+1

2 be the Poisson

kernel on Rn.

(a) Use Exercise 6.1.4 withΨ(x) = ∂∂ t

Pt(x)∣∣t=1

to obtain that the operator

f →(∫ ∞

0t∣∣ ∂∂ t(Pt ∗ f )(x)

∣∣2 dt

)1/2

is bounded from Lp(Rn) to Lp(Rn) for 1 < p < ∞.

(b ) Use Exercise 6.1.4 withΨ(x) = ∂kP1(x) to obtain that the operator

f →(∫ ∞

0t|∂k(Pt ∗ f )(x)|2 dt

)1/2


(c) Conclude that the g-function

g( f )(x) =

(∫ ∞

0t|∇x,t(Pt ∗ f )(x)|2 dt

)1/2


6.3.2. Suppose that µ is a finite Borel measure on Rn with compact support that sat-

isfies µ(0) = 0 and |µ(ξ )| ≤C|ξ |−a for some a > 0 and all ξ = 0. Define measures

µ j by setting µ j(ξ ) = µ(2− jξ ). Show that the operator

Tµ( f )(x) = ∑j∈Z

( f ∗µ j)(x)

is bounded from Lp(Rn) to Lp(Rn) for all 1 0 and all ξ = 0. Show that the

maximal function

Mµ( f )(x) = supj∈Z

∣∣∣∣∫

Rnf (x−2 jy)dµ(y)

∣∣∣∣

is bounded from Lp(Rn) to Lp(Rn) for all 1 < p < ∞.

(b) Let µ be the surface measure on the sphere Sn−1 when n≥ 2. Conclude that the

dyadic spherical maximal function Mµ is bounded on Lp(Rn) for all 1 < p < ∞.[Hint: Part (a): Pick a C ∞

0 function ϕ on Rn with ϕ(0) = 1. Then the measure

σ = µ− µ(0)ϕ satisfies the hypotheses of Corollary 6.3.5. Since,

Mµ( f )(x)≤(∑

j

|(σ j ∗ f )(x)|2)1/2

+ |µ(0)|M( f )(x) ,

it follows that Mµ is bounded on Lp(Rn) whenever 1 < p < ∞. Part (b): If µ = dσ

is surface measure on Sn−1, then |dσ(ξ )| ≤C|ξ |− n−12 (Appendices B.4 and B.7).

]

6.3.4. Let Ω be in Lq(Sn−1) for some 1 < q < ∞ and define the absolutely continu-

ous measure

dµ(x) =Ω(x/|x|)|x|n χ1<|x|≤2 dx .

Show that for all a < 1/q′ we have that |µ(ξ )| ≤C|ξ |−a. Under the additional hy-

pothesis that Ω has mean value zero, conclude that the singular integral operator

TΩ ( f )(x) = p.v.

∫

Rn

Ω(y/|y|)|y|n f (x− y)dy =∑

j

f ∗µ j

is Lp bounded for all 1 < p < ∞. This provides an alternative proof of Theorem

5.2.10 under the hypothesis that Ω ∈ Lq(Sn−1).

6.3.5. For a continuous function F on R define

u(F)(x) =

(∫ ∞

0|F(x+ t)+F(x− t)−2F(x)|2 dt

t3

) 12

.

Given f ∈ L1loc(R) we denote by Ff the indefinite integral of f , that is,

Ff (x) =∫ x

0f (t)dt .

Prove that for all 1< p<∞ there exist constants cp and Cp such that for all functions

f ∈ Lp(R) we have

cp

∥∥ f∥∥

Lp ≤∥∥u(Ff )

∥∥Lp ≤Cp

∥∥ f∥∥

Lp .

[Hint: Let ϕ = χ[−1,0]− χ[0,1]. Then

(ϕt ∗ f )(x) =1

t

(Ff (x+ t)+Ff (x− t)−2Ff (x)

)

and the double inequality follows from parts (b) and (c) of Exercise 6.1.4.]

6.3.6. Let m be a bounded function on Rn that is C 1 in a neighborhood of zero, it

satisfies m(0) = 1 and |m(ξ )| ≤ B|ξ |−ε for all ξ = 0, for some B,ε > 0. Define an

operator Tt by setting Tt( f )(ξ ) = f (ξ )m(tξ ). Show that the maximal operator

supN>0

(1

N

∫ N

0

∣∣Tt( f )(x)∣∣2 dt

)12

maps L2(Rn) to itself.

6.4 Haar System, Conditional Expectation, and Martingales 463

[Hint: Majorize this maximal operator by a constant multiple of the sum

M( f )(x)+(∫ ∞

0|Tt( f )(x)− ( f ∗ϕt)(x)|2

dt

t

) 12,

where ϕ is a C ∞0 function such that ϕ(0) = 1.

]

6.3.7. ([150]) Let 0 < β < 1 and p0 = (1−β/2)−1. Suppose that f j j∈Z are L2

functions on the real line with norm at most 1. Assume that each f j is supported in

interval of length 1 and that the orthogonality relation∣∣⟨ f j | fk〉

∣∣ ≤ (1+ | j− k|)−βholds for all j,k ∈ Z.

(a) Let I Z be such that for all j ∈ I the functions f j are supported in a fixed

interval of length 3. Show that for all p satisfying 0 < p≤ 2 there is Cp,β < ∞ such

that ∥∥∥∑j∈I

ε j f j

∥∥∥Lp≤Cp,β |I|1−

β2

whenever ε j are complex numbers with |ε j| ≤ 1.

(b) Under the same hypothesis as in part (a), prove that for all 0 < p < p0 there is a

constant C′p,β < ∞ such that

∥∥∥∑j∈I

c j f j

∥∥∥Lp≤C′p,β

(∑j∈Z

|c j|p) 1

p

for all complex-valued sequences c j j in ℓp.

(c) Derive the conclusion of part (b) without the assumption that the f j are sup-

ported in a fixed interval of length 3.[Hint: Part (a): Pass from Lp to L2 and use the hypothesis. Part (b): Assume

∑ j∈Z |c j|p = 1. For each k = 0,1, . . . , set Ik = j ∈ Z : 2−k−1 < |c j| ≤ 2−k. Write∥∥∑ j∈Z c j f j

∥∥Lp ≤∑∞k=0 2−k

∥∥∑ j∈Ik(c j2

k) f j

∥∥Lp , use part (b), Holder’s inequality, and

the fact that ∑∞k=0 2−kp|Ik| ≤ 2p. Part (c): Write ∑ j∈Z c j f j = ∑m∈Z Fm, where Fm is

the sum of c j f j over all j such that the support of f j meets the interval [m,m+ 1].These Fm’s are supported in [m−1,m+2] and are almost orthogonal.

]

6.4 The Haar System, Conditional Expectation, and Martingales

There is a very strong connection between the Littlewood–Paley operators and cer-

tain notions from probability, such as conditional expectation and martingale differ-

ence operators. The conditional expectation we are concerned with is with respect

to the increasing σ -algebra of all dyadic cubes on Rn.


6.4.1 Conditional Expectation and Dyadic Martingale Differences

We recall the definition of dyadic cubes.

Definition 6.4.1. A dyadic interval in R is an interval of the form

[m2−k,(m+1)2−k

)

where m,k are integers. A dyadic cube in Rn is a product of dyadic intervals of the

same length. That is, a dyadic cube is a set of the form

n

∏j=1

[m j2

−k,(m j +1)2−k)

for some integers m1, . . . ,mn,k.

We defined dyadic intervals to be closed on the left and open on the right, so that

different dyadic intervals of the same length are always disjoint sets.

Given a cube Q in Rn we denote by |Q| its Lebesgue measure and by ℓ(Q) its

side length. We clearly have |Q|= ℓ(Q)n. We introduce some more notation.

Definition 6.4.2. For k ∈ Z we denote by Dk the set of all dyadic cubes in Rn whose

side length is 2−k. We also denote by D the set of all dyadic cubes in Rn. Then we

have

D =⋃

k∈Z

Dk ,

and moreover, the σ -algebra σ(Dk) of measurable subsets of Rn formed by count-

able unions and complements of elements of Dk is increasing as k increases.

We observe the fundamental property of dyadic cubes, which clearly justifies

their usefulness. Any two dyadic intervals of the same side length either are disjoint

or coincide. Moreover, either two given dyadic intervals are disjoint, or one contains

the other. Similarly, either two dyadic cubes are disjoint, or one contains the other.

Definition 6.4.3. Given a locally integrable function f on Rn, we denote by

AvgQ

f =1

|Q|

∫

Qf (t)dt

the average of f over a cube Q.

The conditional expectation of a locally integrable function f on Rn with respect

to the increasing family of σ -algebras σ(Dk) generated by Dk is defined as

Ek( f )(x) = ∑Q∈Dk

(AvgQ

f )χQ(x),

for all k∈Z. We also define the dyadic martingale difference operator Dk as follows:

Dk( f ) = Ek( f )−Ek−1( f ),

also for k ∈ Z.

Next we introduce the family of Haar functions.

Definition 6.4.4. For a dyadic interval I = [m2−k,(m + 1)2−k) we define IL =[m2−k,(m+ 1

2)2−k) and IR = [(m+ 1

2)2−k,(m+1)2−k) to be the left and right parts

of I, respectively. The function

hI(x) = |I|−12 χIL −|I|−

12 χIR

is called the Haar function associated with the interval I.

We remark that Haar functions are constructed in such a way that they have L2

norm equal to 1. Moreover, the Haar functions have the following fundamental or-

thogonality property:

∫

RhI(x)hI′(x)dx =

0 when I = I′,

1 when I = I′.(6.4.1)

To see this, observe that the Haar functions have L2 norm equal to 1 by construction.

Moreover, if I = I′, then I and I′ must have different lengths, say we have |I′|< |I|.If I and I′ are not disjoint, then I′ is contained either in the left or in the right half of

I, on either of which hI is constant. Thus (6.4.1) follows.

We recall the notation

⟨f ,g

⟩=

∫

Rf (x)g(x)dx

valid for square integrable functions. Under this notation, (6.4.1) can be rewritten as⟨hI ,hI′

⟩= δI,I′ , where the latter is 1 when I = I′ and zero otherwise.

6.4.2 Relation Between Dyadic Martingale Differences and Haar

Functions

We have the following result relating the Haar functions to the dyadic martingale

difference operators in dimension one.

Proposition 6.4.5. For every locally integrable function f on R and for all k ∈Z we

have the identity

Dk( f ) = ∑I∈Dk−1

⟨f ,hI

⟩hI (6.4.2)


and also ∥∥Dk( f )∥∥2

L2 = ∑I∈Dk−1

∣∣⟨ f ,hI

⟩∣∣2. (6.4.3)

Proof. We observe that every interval J in Dk is either an IL or an IR for some unique

I ∈Dk−1. Thus we can write

Ek( f ) = ∑J∈Dk

(AvgJ

f )χJ

= ∑I∈Dk−1

[(2

|I|

∫

IL

f (t)dt

)χIL +

(2

|I|

∫

IR

f (t)dt

)χIR

].

(6.4.4)

But we also have

Ek−1( f ) = ∑I∈Dk−1

(AvgI

f )χI

= ∑I∈Dk−1

(1

|I|

∫

IL

f (t)dt +1

|I|

∫

IR

f (t)dt

)(χIL + χIR

).

(6.4.5)

Now taking the difference between (6.4.4) and (6.4.5) we obtain

Dk( f ) = ∑I∈Dk−1

[(1

|I|

∫

IL

f (t)dt

)χIL −

(1

|I|

∫

IR

f (t)dt

)χIL

+

(1

|I|

∫

IR

f (t)dt

)χIR −

(1

|I|

∫

IL

f (t)dt

)χIR

],

which is easily checked to be equal to

∑I∈Dk−1

(∫

If (t)hI(t)dt

)hI = ∑

I∈Dk−1

⟨f ,hI

⟩hI .

Finally, (6.4.3) is a consequence of (6.4.1).

Theorem 6.4.6. Every function f ∈ L2(Rn) can be written as

f = ∑k∈Z

Dk( f ) , (6.4.6)

where the series converges almost everywhere and in L2. We also have

∥∥ f∥∥2

L2(Rn)= ∑

k∈Z

∥∥Dk( f )∥∥2

L2(Rn). (6.4.7)

Moreover, when n = 1 we have the representation

f = ∑I∈D

⟨f ,hI

⟩hI , (6.4.8)


where the sum converges a.e. and in L2 and also

∥∥ f∥∥2

L2(R)= ∑

I∈D

∣∣⟨ f ,hI

⟩∣∣2 . (6.4.9)

Proof. In view of the Lebesgue differentiation theorem, Corollary 2.1.16, given a

function f ∈ L2(Rn) there is a set N f of measure zero on Rn such that for all x ∈Rn \N f we have that

AvgQ j

f → f (x)

whenever Q j is a sequence of decreasing cubes such that⋂

j Q j = x. Given x

in Rn \N f there exists a unique sequence of dyadic cubes Q j(x) ∈ D j such that⋂∞j=0 Q j(x) = x. Then for all x ∈ Rn \N f we have

limj→∞

E j( f )(x) = limj→∞

∑Q∈D j

(AvgQ

f )χQ(x) = limj→∞

AvgQ j(x)

f = f (x) .

From this we conclude that E j( f )→ f a.e. as j → ∞. We also observe that since

|E j( f )| ≤Mc( f ), where Mc denotes the uncentered maximal function with respect

to cubes, we have that |E j( f )− f | ≤ 2Mc( f ), which allows us to obtain from the

Lebesgue dominated convergence theorem that E j( f )→ f in L2 as j→ ∞.

Next we study convergence of E j( f ) as j →−∞. For a given x ∈ Rn and Q j(x)as before we have that

|E j( f )(x)|=∣∣ Avg

Q j(x)

f∣∣≤

(1

|Q j(x)|

∫

Q j(x)| f (t)|2 dt

)12

≤ 2jn2

∥∥ f∥∥

L2 ,

which tends to zero as j →−∞, since the side length of each Q j(x) is 2− j. Since

|E j( f )| ≤ Mc( f ), the Lebesgue dominated convergence theorem allows us to con-

clude that E j( f )→ 0 in L2 as j→−∞. To obtain the conclusion asserted in (6.4.6)

we simply observe that

N

∑k=M

Dk( f ) = EN( f )−EM−1( f )→ f

in L2 and almost everywhere as N → ∞ and M →−∞.

To prove (6.4.7) we first observe that we can rewrite Dk( f ) as

Dk( f ) = ∑Q∈Dk

(AvgQ

f )χQ− ∑R∈Dk−1

(AvgR

f )χR

= ∑R∈Dk−1

[∑

Q∈Dk

Q⊆R

(AvgQ

f )χQ− (AvgR

f )χR

]


= ∑R∈Dk−1

[∑

Q∈Dk

Q⊆R

(AvgQ

f )χQ−1

2n ∑Q∈Dk

Q⊆R

(AvgQ

f )χR

]

= ∑R∈Dk−1

∑Q∈Dk

Q⊆R

(AvgQ

f )(χQ−2−nχR

). (6.4.10)

Using this identity we obtain that for given integers k′ > k we have

∫

RnDk( f )(x)Dk′( f )(x)dx

= ∑R∈Dk−1

∑Q∈DkQ⊆R

(AvgQ

f ) ∑R′∈Dk′−1

∑Q′∈Dk′Q′⊆R′

(AvgQ′

f )∫ (

χQ−2−nχR

)(χQ′ −2−nχR′

)dx .

Since k′ > k, the last integral may be nonzero only when R′ R. If this is the case,

then R′ ⊆ QR′ for some dyadic cube QR′ ∈Dk with QR′ R. See Figure 6.1.

Fig. 6.1 Picture of the cubes

R, R′, and QR′ .

Then the function χQ′ − 2−nχR′ is supported in the cube QR′ and the function

χQ−2−nχR is constant on any dyadic subcube Q of R (of half its side length) and in

particular is constant on QR′ . Then

∑Q′∈Dk′Q′⊆R′

(Avg

Q′f)∫

QR′χQ′ −2−nχR′ dx = ∑

Q′∈Dk′Q′⊆R′

(Avg

Q′f)(|Q′|−2−n|R′|

)= 0 ,

since |R′| = 2n|Q′|. We conclude that⟨Dk( f ),Dk′( f )

⟩= 0 whenever k = k′, from

which we easily derive (6.4.7).

Now observe that (6.4.8) is a direct consequence of (6.4.2), and (6.4.9) is a direct

consequence of (6.4.3).

6.4.3 The Dyadic Martingale Square Function

As a consequence of identity (6.4.7), proved in the previous subsection, we obtain

that ∥∥∥(∑k∈Z

|Dk( f )|2) 1

2∥∥∥

L2(Rn)=

∥∥ f∥∥

L2(Rn), (6.4.11)

which says that the dyadic martingale square function

S( f ) =(∑k∈Z

|Dk( f )|2) 1

2

is L2 bounded. It is natural to ask whether there exist Lp analogues of this result, and

this is the purpose of the following theorem.

Theorem 6.4.7. For any 1 < p < ∞ there exists a constant cp,n such that for every

function f in Lp(Rn) we have

1

cp′,n

∥∥ f∥∥

Lp(Rn)≤

∥∥S( f )∥∥

Lp(Rn)≤ cp,n

∥∥ f∥∥

Lp(Rn). (6.4.12)

The lower inequality subsumes the fact that if∥∥S( f )

∥∥Lp(Rn)

< ∞, then f must be an

Lp function.

Proof. Let r j j be the Rademacher functions (see Appendix C.1) enumerated in

such a way that their index set is the set of integers. We rewrite the upper estimate

in (6.4.12) as

∫ 1

0

∫

Rn

∣∣∣∑k∈Z

rk(ω)Dk( f )(x)∣∣∣

p

dxdω ≤Cpp

∥∥ f∥∥p

Lp . (6.4.13)

We prove a stronger estimate than (6.4.13), namely that for all ω ∈ [0,1] we have

∫

Rn

∣∣∣Tω( f )(x)∣∣∣

p

dx≤Cpp

∥∥ f∥∥p

Lp , (6.4.14)

where

Tω( f )(x) = ∑k∈Z

rk(ω)Dk( f )(x) .

In view of the L2 estimate (6.4.11), we have that the operator Tω is L2 bounded with

norm 1. We show that Tω is weak type (1,1).To show that Tω is of weak type (1,1) we fix a function f ∈ L1 and α > 0. We

apply the Calderon–Zygmund decomposition (Theorem 5.3.1) to f at height α to

write

f = g+b, b =∑j

(f −Avg

Q j

f)χQ j

,

where Q j are dyadic cubes that satisfy ∑ j |Q j| ≤ 1α

∥∥ f∥∥

L1 and g has L2 norm at

most (2nα∥∥ f

∥∥L1)

12 ; see (5.3.1). To achieve this decomposition, we apply the proof

of Theorem 5.3.1 starting with a dyadic mesh of large cubes such that |Q| ≥ 1α

∥∥ f∥∥

L1

for all Q in the mesh. Then we subdivide each Q in the mesh by halving each side,

and we select those cubes for which the average of f over them is bigger than α (and

thus at most 2nα). Since the original mesh consists of dyadic cubes, the stopping-

time argument of Theorem 5.3.1 ensures that each selected cube is dyadic.

We observe (and this is the key observation) that Tω(b) is supported in⋃

j Q j. To

see this, we use identity (6.4.10) to write Tω(b) as

∑j

[∑k

rk(ω) ∑R∈Dk−1

∑Q∈DkQ⊆R

AvgQ

[( f −AvgQ j

f )χQ j](χQ−2−nχR

)]. (6.4.15)

We consider the following three cases for the cubes Q that appear in the inner sum in

(6.4.15): (i) Q j ⊆Q, (ii) Q j∩Q = /0, and (iii) Q Q j. It is simple to see that in cases

(i) and (ii) we have AvgQ[( f −AvgQ jf )χQ j

] = 0. Therefore the inner sum in (6.4.15)

is taken over all Q that satisfy Q Q j. But then we must have that the unique dyadic

parent R of Q is also contained in Q j. It follows that the expression inside the square

brackets in (6.4.15) is supported in R and therefore in Q j. We conclude that Tω(b)is supported in

⋃j Q j. Using Exercise 5.3.5(a) we obtain that Tω is weak type (1,1)

with norm at most

α∣∣|Tω(g)|> α

2∣∣+α

∣∣⋃j Q j

∣∣‖ f‖L1

≤α4α−2‖g‖2

L2 +‖ f‖L1

‖ f‖L1

≤ 2n+2 +1 .

We have now established that Tω is weak type (1,1). Since Tω is L2 bounded with

norm 1, it follows by interpolation that Tω is Lp bounded for all 1 2 follows by duality. (Note that the

operators Dk and Ek are self-transpose.) We conclude the validity of (6.4.14), which

implies that of (6.4.13). As observed, this is equivalent to the upper estimate in

(6.4.12).

Finally, we notice that the lower estimate in (6.4.12) is a consequence of the

upper estimate as in the case of the Littlewood–Paley operators ∆ j. Indeed, we need

to observe that in view of (6.4.6) we have

∣∣⟨ f ,g⟩∣∣ =

∣∣⟨∑k

Dk( f ),∑k′

Dk′(g)⟩∣∣

=∣∣∣∑

k

∑k′

⟨Dk( f ),Dk′(g)

⟩∣∣∣

=∣∣∣∑

k

⟨Dk( f ),Dk(g)

⟩∣∣∣ [Exercise 6.4.6(a)]

≤∫

Rn∑k

|Dk( f )(x)| |Dk(g)(x)|dx

≤∫

RnS( f )(x)S(g)(x)dx (Cauchy–Schwarz inequality)

≤∥∥S( f )

∥∥Lp

∥∥S(g)∥∥

Lp′ (Holder’s inequality)

≤∥∥S( f )

∥∥Lp cp′,n

∥∥g∥∥

Lp′ .

Taking the supremum over all functions g on Rn with Lp′ norm at most 1, we obtain

that f gives rise to a bounded linear functional on Lp′ . It follows by the Riesz repre-

sentation theorem that f must be an Lp function that satisfies the lower estimate in

(6.4.12).

6.4.4 Almost Orthogonality Between the Littlewood–Paley

Operators and the Dyadic Martingale Difference Operators

Next, we discuss connections between the Littlewood–Paley operators ∆ j and the

dyadic martingale difference operators Dk. It turns out that these operators are al-

most orthogonal in the sense that the L2 operator norm of the composition Dk∆ j

decays exponentially as the indices j and k get farther away from each other.

For the purposes of the next theorem we define the Littlewood–Paley operators

∆ j as convolution operators with the functionΨ2− j , where

Ψ(ξ ) = Φ(ξ )− Φ(2ξ )

and Φ is a fixed radial Schwartz function whose Fourier transform Φ is real-valued,

supported in the ball |ξ | < 2, and equal to 1 on the ball |ξ | < 1. In this case we

clearly have the identity

∑j∈Z

Ψ(2− jξ ) = 1, ξ = 0 .

Then we have the following theorem.

Theorem 6.4.8. There exists a constant C such that for every k, j in Z the following

estimate on the operator norm of Dk∆ j : L2(Rn)→ L2(Rn) is valid:

∥∥Dk∆ j

∥∥L2(Rn)→L2(Rn)

=∥∥∆ jDk

∥∥L2(Rn)→L2(Rn)

≤C 2−12 | j−k|. (6.4.16)

Proof. SinceΨ is a radial function, it follows that ∆ j is equal to its transpose oper-

ator on L2. Moreover, the operator Dk is also equal to its transpose. Thus

(Dk∆ j)t = ∆ jDk

and it therefore suffices to prove only that

∥∥Dk∆ j

∥∥L2→L2 ≤C2−

12 | j−k| . (6.4.17)


By a simple dilation argument it suffices to prove (6.4.17) when k = 0. In this

case we have the estimate

∥∥D0∆ j

∥∥L2→L2 =

∥∥E0∆ j−E−1∆ j

∥∥L2→L2

≤∥∥E0∆ j−∆ j

∥∥L2→L2 +

∥∥E−1∆ j−∆ j

∥∥L2→L2 ,

and since the Dk’s and ∆ j’s are self-transposes, we have

∥∥D0∆ j

∥∥L2→L2 =

∥∥∆ jD0

∥∥L2→L2 =

∥∥∆ jE0−∆ jE−1

∥∥L2→L2

≤∥∥∆ jE0−E0

∥∥L2→L2 +

∥∥∆ jE−1−E0

∥∥L2→L2 .

Estimate (6.4.17) when k = 0 will be a consequence of the pair of inequalities

∥∥E0∆ j−∆ j

∥∥L2→L2 +

∥∥E−1∆ j−∆ j

∥∥L2→L2 ≤C 2

j2 for j ≤ 0, (6.4.18)

∥∥∆ jE0−E0

∥∥L2→L2 +

∥∥∆ jE−1−E0

∥∥L2→L2 ≤C 2−

12 j for j ≥ 0. (6.4.19)

We start by proving (6.4.18). We consider only the term E0∆ j−∆ j, since the term

E−1∆ j−∆ j is similar. Let f ∈ L2(Rn). Then

∥∥E0∆ j( f )−∆ j( f )∥∥2

L2

= ∑Q∈D0

∥∥ f ∗Ψ2− j −AvgQ

( f ∗Ψ2− j)∥∥2

L2(Q)

≤ ∑Q∈D0

∫

Q

∫

Q|( f ∗Ψ2− j)(x)− ( f ∗Ψ2− j)(t)|2 dt dx

≤ 3 ∑Q∈D0

∫

Q

∫

Q

(∫

5√

nQ| f (y)||Ψ2− j(x− y)|dy

)2

dt dx

+3 ∑Q∈D0

∫

Q

∫

Q

(∫

5√

nQ| f (y)||Ψ2− j(t− y)|dy

)2

dt dx

+3 ∑Q∈D0

∫

Q

∫

Q

(∫

(5√

nQ)c| f (y)|2 jn+ j|∇Ψ(2 j(ξx,t − y))|dy

)2

dt dx,

where ξx,t lies on the line segment joining x and t. Applying the Cauchy-Schwarz

inequality to the first two terms, we see that the last expression is bounded by

C2 jn ∑Q∈D0

∫

5√

nQ| f (y)|2 dy+CM22 j ∑

Q∈D0

∫

Q

(∫

Rn

2 jn| f (y)|dy

(1+2 j|x− y|)M

)2

dx ,

which is clearly controlled by C(2 jn+22 j)‖ f‖2L2 ≤ 2C2 j‖ f‖2

L2 . This proves (6.4.18).

We now turn to the proof of (6.4.19). We set S j = ∑k≤ j∆k. Since ∆ j is the differ-

ence of two S j’s, it suffices to prove (6.4.19), where ∆ j is replaced by S j. We work

only with the term S jE0−E0, since the other term can be treated similarly. We have


∥∥S jE0( f )−E0( f )∥∥2

L2 =∥∥∥ ∑

Q∈D0

(AvgQ

f )(Φ2− j ∗χQ− χQ)∥∥∥

2

L2

≤ 2

∥∥∥ ∑Q∈D0

(AvgQ

f )(Φ2− j ∗χQ− χQ)χ5√

nQ

∥∥∥2

L2

+2

∥∥∥ ∑Q∈D0

(AvgQ

f )(Φ2− j ∗χQ)χ(5√

nQ)c

∥∥∥2

L2.

Since the functions appearing inside the sum in the first term have supports with

bounded overlap, we obtain

∥∥∥ ∑Q∈D0

(AvgQ

f )(Φ2− j ∗χQ− χQ)χ5√

nQ

∥∥∥2

L2≤C ∑

Q∈D0

(AvgQ

| f |)2∥∥Φ2− j ∗χQ− χQ

∥∥2

L2 ,

and the crucial observation is that

∥∥Φ2− j ∗χQ− χQ

∥∥2

L2 ≤C 2− j,

a consequence of Plancherel’s identity and the fact that |1− Φ(2− jξ )| ≤ χ|ξ |≥2 j .

Putting these observations together, we deduce

∥∥∥ ∑Q∈D0

(AvgQ

f )(Φ2− j ∗χQ− χQ)χ3Q

∥∥∥2

L2≤C ∑

Q∈D0

(AvgQ

| f |)22− j ≤C 2− j∥∥ f

∥∥2

L2 ,

and the required conclusion will be proved if we can show that

∥∥∥ ∑Q∈D0

(AvgQ

f )(Φ2− j ∗χQ)χ(3Q)c

∥∥∥2

L2≤C2− j

∥∥ f∥∥2

L2 . (6.4.20)

We prove (6.4.20) by using an estimate based purely on size. Let cQ be the center of

the dyadic cube Q. For x /∈ 3Q we have the estimate

|(Φ2− j ∗χQ)(x)| ≤CM2 jn

(1+2 j|x− cQ|)M≤ CM2 jn

(1+2 j)M/2

1

(1+ |x− cQ|)M/2,

since both 2 j ≥ 1, and |x−cQ| ≥ 1. We now control the left-hand side of (6.4.20) by

2 j(2n−M) ∑Q∈D0

∑Q′∈D0

(AvgQ

| f |)(AvgQ′| f |)

∫

Rn

CM dx

(1+|x−cQ|)M2 (1+|x−cQ′ |)

M2

≤ 2 j(2n−M) ∑Q∈D0

∑Q′∈D0

(AvgQ

| f |)(AvgQ′| f |)

(1+ |cQ− cQ′ |)M4

∫

Rn

CM dx

(1+|x−cQ|)M4 (1+|x−cQ′ |)

M4

≤ 2 j(2n−M) ∑Q∈D0

∑Q′∈D0

CM

(1+ |cQ− cQ′ |)M4

(∫

Q| f (y)|2 dy+

∫

Q′| f (y)|2 dy

)

≤CM2 j(2n−M) ∑Q∈D0

∫

Q| f (y)|2 dy

=CM2 j(2n−M)∥∥ f

∥∥2

L2 .

By taking M large enough, we obtain (6.4.20) and thus (6.4.19).

Exercises

6.4.1. (a) Prove that no dyadic cube in Rn contains the point 0 in its interior.

(b) Prove that every interval [a,b] is contained in the union of three dyadic intervals

of length less than b−a.

(c) Prove that every cube of length l in Rn is contained in the union of 3n dyadic

cubes, each having length less than l.

6.4.2. Let k ∈ Z. Show that the set [m2−k,(m+ s)2−k) is a dyadic interval if and

only if s = 2p for some p ∈ Z and m is an integer multiple of s.

6.4.3. Given a cube Q in Rn of side length ℓ(Q) ≤ 2k−1 for some integer k, prove

that there is a dyadic cube DQ of side length 2k such that Q σ +DQ for some

σ = (σ1, . . . ,σn), where σ j ∈ 0,1/3,−1/3.6.4.4. Show that the martingale maximal function f → supk∈Z |Ek( f )| is weak type

(1,1) with constant at most 1.[Hint: Use Exercise 2.1.12.

]

6.4.5. (a) Show that EN( f )→ f a.e. as N → ∞ for all f ∈ L1loc(R

n).(b) Prove that EN( f )→ f in Lp as N → ∞ for all f ∈ Lp(Rn) whenever 1 < p < ∞.

6.4.6. (a) Let k,k′ ∈Z be such that k = k′. Show that for functions f and g in L2(Rn)we have ⟨

Dk( f ),Dk′(g)⟩= 0 .

(b) Conclude that for functions f j in L2(Rn) we have

∥∥∥∑j∈Z

D j( f j)∥∥∥

L2(Rn)=

(∑j∈Z

∥∥D j( f j)∥∥2

L2(Rn)

) 12.

(c) Let ∆ j and C be as in the statement of Theorem 6.4.8. Show that for any r ∈ Z

we have ∥∥∥∑j∈Z

D j∆ j+rD j

∥∥∥L2(Rn)→L2(Rn)

≤C 2−12 |r| .

6.4.7. ([133]) Let D j, ∆ j be as in Theorem 6.4.8.

(a) Prove that the operator

Vr = ∑j∈Z

D j∆ j+r

is bounded from L2(Rn) to itself with norm at most a multiple of 2−12 |r|.

6.5 The Spherical Maximal Function 475

(b) Show that Vr is Lp(Rn) bounded for all 1 0 such that Vr is

bounded on Lp(Rn) with norm at most a multiple of 2−cp |r|.[Hint: Part (a): Write ∆ j = ∆ j∆ j, where ∆ j is another family of Littlewood–Paley

operators and use Exercise 6.4.6 (b). Part (b): Use duality and (6.1.21).]

6.5 The Spherical Maximal Function

In this section we discuss yet another consequence of the Littlewood–Paley theory,

the boundedness of the spherical maximal operator.

6.5.1 Introduction of the Spherical Maximal Function

We denote throughout this section by dσ the normalized Lebesgue measure on the

sphere Sn−1. For f in Lp(Rn), 1≤ p≤ ∞, we define the maximal operator

M ( f )(x) = supt>0

∣∣∣∣∫

Sn−1f (x− tθ)dσ(θ)

∣∣∣∣ (6.5.1)

and we observe that by Minkowski’s integral inequality each expression inside the

supremum in (6.5.1) is well defined for f ∈ Lp for almost all x ∈ Rn. The operator

M is called the spherical maximal function. It is unclear at this point for which

functions f we have M ( f ) < ∞ a.e. and for which values of p < ∞ the maximal

inequality ∥∥M ( f )∥∥

Lp(Rn)≤Cp

∥∥ f∥∥

Lp(Rn)(6.5.2)

holds for all functions f ∈ Lp(Rn).Spherical averages often make their appearance as solutions of partial differential

equations. For instance, the spherical average

u(x, t) =1

4π

∫

S2t f (x− ty)dσ(y) (6.5.3)

is a solution of the wave equation

∆x(u)(x, t) =∂ 2u

∂ t2(x, t) ,

u(x,0) = 0 ,

∂u

∂ t(x,0) = f (x) ,

in R3. The introduction of the spherical maximal function is motivated by the fact

that the related spherical average

u(x, t) =1

4π

∫

S2f (x− ty)dσ(y) (6.5.4)

solves Darboux’s equation

∆x(u)(x, t) =∂ 2u

∂ t2(x, t)+

2

t

∂u

∂ t(x, t) ,

u(x,0) = f (x) ,

∂u

∂ t(x,0) = 0 ,

in R3. It is rather remarkable that the Fourier transform can be used to study almost

everywhere convergence for several kinds of maximal averaging operators such as

the spherical averages in (6.5.4). This is achieved via the boundedness of the cor-

responding maximal operator; the maximal operator controlling the averages over

Sn−1 is given in (6.5.1).

Before we begin the analysis of the spherical maximal function, we recall that

dσ(ξ ) =2π

|ξ | n−22

J n−22(2π|ξ |) ,

as shown in Appendix B.4. Using the estimates in Appendices B.6 and B.7 and the

identityd

dtJν(t) =

1

2(Jν−1(t)− Jν+1(t))

derived in Appendix B.2, we deduce the crucial estimate

|dσ(ξ )|+ |∇dσ(ξ )| ≤ Cn

(1+ |ξ |) n−12

. (6.5.5)

Theorem 6.5.1. Let n≥ 3. For each nn−1

< p≤ ∞, there is a constant Cp such that

∥∥M ( f )∥∥

Lp(Rn)≤Cp

∥∥ f∥∥

Lp(Rn)(6.5.6)

holds for all f in Lp(Rn). Consequently, for all nn−1

< p ≤ ∞ and f ∈ Lp(Rn) we

have

limt→0

1

ωn−1

∫

Sn−1f (x− tθ)dσ(θ) = f (x) (6.5.7)

for almost all x ∈ Rn. Here we set ωn−1 = |Sn−1|.

The proof of this theorem is given in the rest of this section. Before we present the

proof we explain the validity of (6.5.7). Clearly this assertion is valid for functions

f ∈S (Rn). Using inequality (6.5.6) and Theorem 2.1.14 we obtain that (6.5.7) holds

for all functions in f ∈ Lp(Rn).

We now focus on (6.5.6). Define m(ξ ) = dσ(ξ ) and notice that m(ξ ) is a C ∞

function. To study the maximal multiplier operator

supt>0

∣∣( f (ξ )m(tξ ))∨∣∣

we decompose the multiplier m(ξ ) into radial pieces as follows: We fix a radial C ∞

function ϕ0 in Rn such that ϕ0(ξ ) = 1 when |ξ | ≤ 1 and ϕ0(ξ ) = 0 when |ξ | ≥ 2.

For j ≥ 1 we let

ϕ j(ξ ) = ϕ0(2− jξ )−ϕ0(2

1− jξ ) (6.5.8)

and we observe that ϕ j(ξ ) is localized near |ξ | ≈ 2 j. Then we have

∞

∑j=0

ϕ j = 1 .

Set m j = ϕ jm for all j ≥ 0. The m j’s are C ∞0 functions that satisfy

m =∞

∑j=0

m j .

Also, the following estimate is valid:

M ( f )≤∞

∑j=0

M j( f ) ,

where

M j( f )(x) = supt>0

∣∣( f (ξ )m j(tξ ))∨(x)

∣∣ .

Since the function m0 is C ∞0 , we have that M0 maps Lp to itself for all 1 < p ≤ ∞.

(See Exercise 6.5.1.)

We define g-functions associated with m j as follows:

G j( f )(x) =

(∫ ∞

0|A j,t( f )(x)|2 dt

t

)12

,

where A j,t( f )(x) =(

f (ξ )m j(tξ ))∨(x).

6.5.2 The First Key Lemma

We have the following lemma:

Lemma 6.5.2. There is a constant C = C(n) < ∞ such that for any j ≥ 1 we have

the estimate ∥∥M j( f )∥∥

L2 ≤C2(12− n−1

2 ) j∥∥ f

∥∥L2

for all functions f in L2(Rn).

Proof. We define a function

m j(ξ ) = ξ ·∇m j(ξ ) ,

we let A j,t( f )(x) =(

f (ξ ) m j(tξ ))∨

(x), and we let

G j( f )(x) =

(∫ ∞

0|A j,t( f )(x)|2 dt

t

)12

be the associated g-function. For f ∈ L2(Rn), the identity

sdA j,s

ds( f ) = A j,s( f )

is clearly valid for all j and s. Since A j,s( f ) = f ∗ (m∨j )s and m∨j has integral zero for

j ≥ 1 (here (m∨j )s(x) = s−nm∨j (s−1x)), it follows from Corollary 2.1.19 that

lims→0

A j,s( f )(x) = 0

for all x ∈ Rn \E f , where E f is some set of Lebesgue measure zero. By the funda-

mental theorem of calculus for x ∈ Rn \E f we deduce that

(A j,t( f )(x))2 =∫ t

0

d

ds(A j,s( f )(x))2 ds

= 2

∫ t

0A j,s( f )(x)s

dA j,s

ds( f )(x)

ds

s

= 2

∫ t

0A j,s( f )(x)A j,s( f )(x)

ds

s,


∣∣A j,t( f )(x)∣∣2 ≤ 2

∫ ∞

0

∣∣A j,s( f )(x)∣∣ ∣∣A j,s( f )(x)

∣∣ ds

s. (6.5.9)

Taking the supremum over all t > 0 on the left-hand side in (6.5.9) and integrating

over Rn, we obtain the estimate

∥∥M j( f )∥∥2

L2 ≤ 2

∫

Rn

∫ ∞

0

∣∣A j,s( f )(x)∣∣ ∣∣A j,s( f )(x)

∣∣ ds

sdx

≤ 2

∫

RnG j( f )(x)G j( f )(x)dx

≤ 2∥∥G j( f )

∥∥L2

∥∥G j( f )∥∥

L2 ,

by applying the Cauchy–Schwarz inequality twice. Next we claim that as a conse-

quence of (6.5.5) we have for some c, c < ∞,

∥∥m j

∥∥L∞≤ c2− j n−1

2 and∥∥m j

∥∥L∞≤ c2 j(1− n−1

2 ) .

Using these facts together with the facts that the functions m j and m j are sup-

ported in the annuli 2 j−1 ≤ |ξ | ≤ 2 j+1, we obtain that the g-functions G j and G j

are L2 bounded with norms at most a constant multiple of the quantities 2− j n−12 and

2 j(1− n−12 ), respectively; see Exercise 6.5.2. Note that since n≥ 3, both exponents are

negative. We conclude that

∥∥M j( f )∥∥

L2 ≤C2 j( 12− n−1

2 )∥∥ f

∥∥L2 ,

which is what we needed to prove.

6.5.3 The Second Key Lemma

Next we need the following lemma.

Lemma 6.5.3. There exists a constant C =C(n)< ∞ such that for all j ≥ 1 and for

all f in L1(Rn) we have

∥∥M j( f )∥∥

L1,∞ ≤C 2 j∥∥ f

∥∥L1 .

Proof. Let K( j) = (ϕ j)∨ ∗dσ =Φ2− j ∗dσ , where Φ is a Schwartz function. Setting

(K( j))t(x) = t−nK( j)(t−1x)

we have that

M j( f ) = supt>0

|(K( j))t ∗ f | . (6.5.10)

The proof of the lemma is based on the estimate:

M j( f )≤C 2 jM( f ) (6.5.11)

and the weak type (1,1) boundedness of the Hardy–Littlewood maximal operator M

(Theorem 2.1.6). To establish (6.5.11), it suffices to show that for any M > n there is

a constant CM < ∞ such that

|K( j)(x)|= |(Φ2− j ∗dσ)(x)| ≤ CM 2 j

(1+ |x|)M. (6.5.12)

Then Theorem 2.1.10 yields (6.5.11) and hence the required conclusion.

Using the fact that Φ is a Schwartz function, we have for every N > 0,

|(Φ2− j ∗dσ)(x)| ≤CN

∫

Sn−1

2n j dσ(y)

(1+2 j|x− y|)N.

We pick an N to depend on M (6.5.12); in fact, any N > M suffices for our purposes.

We split the last integral into the regions

S−1(x) = Sn−1∩y ∈ Rn : 2 j|x− y| ≤ 1

and for r ≥ 0,

Sr(x) = Sn−1∩y ∈ Rn : 2r < 2 j|x− y| ≤ 2r+1 .

The key observation is that whenever B(y,R) is a ball of radius R in Rn centered at

y ∈ Sn−1, then the spherical measure of the set Sn−1 ∩B(y,R) is at most a dimen-

sional constant multiple of Rn−1. This implies that the spherical measure of each

Sr(x) is at most cn2(r+1− j)(n−1), an estimate that is useful only when r ≤ j. Using

this observation, together with the fact that for y ∈ Sr(x) we have |x| ≤ 2r+1− j + 1,

we obtain the following estimate for the expression |(Φ2− j ∗dσ)(x)|:

j

∑r=−1

∫

Sr(x)

CN2n j dσ(y)

(1+2 j|x− y|)N+

∞

∑r= j+1

∫

Sr(x)

CN2n j dσ(y)

(1+2 j|x− y|)N

≤C′N2n j

[ j

∑r=−1

dσ(Sr(x))χB(0,3)(x)

2rN+

∞

∑r= j+1

dσ(Sr(x))χB(0,2r+1− j+1)(x)

2rN

]

≤C′N2n j

[ j

∑r=−1

cn2(r+1− j)(n−1)χB(0,3)(x)

2rN+

∞

∑r= j+1

ωn−1 χB(0,2r+2− j)(x)

2rN

]

≤CN,n

[2 jχB(0,3)(x)+2n j

∞

∑r= j+1

1

2rN

(1+2r+2− j)M

(1+ |x|)M

]

≤C′M,n2 j

(1+ |x|)M

[1+

∞

∑r= j+1

2(r− j)(M−N)

2 j(N+1−n)

]

≤C′′M,n2 j

(1+ |x|)M,

where we used that N > M > n. This establishes (6.5.12).

6.5.4 Completion of the Proof

It remains to combine the previous ingredients to complete the proof of the theorem.

Interpolating between the L2 → L2 and L1 → L1,∞ estimates obtained in Lemmas

6.5.2 and 6.5.3, we obtain

∥∥M j( f )∥∥

Lp(Rn)≤Cp2

( np−(n−1)) j

∥∥ f∥∥

Lp(Rn)

for all 1 nn−1

the series ∑∞j=1 2( n

p−(n−1)) jconverges and we

conclude that M is Lp bounded for these p’s. The boundedness of M on Lp for

p > 2 follows by interpolation between Lq for q < 2 and the estimate M : L∞→ L∞.

Exercises

6.5.1. Let m be in L1(Rn)∩L∞(Rn) that satisfies |m∨(x)| ≤C(1+ |x|)−n−δ for some

δ > 0. Show that the maximal multiplier

Mm( f )(x) = supt>0

∣∣( f (ξ )m(tξ ))∨(x)

∣∣

is Lp bounded for all 1 < p < ∞.

6.5.2. Suppose that the function m is supported in the annulus R≤ |ξ | ≤ 2R and is

bounded by A. Show that the g-function

G( f )(x) =

(∫ ∞

0|(m(tξ ) f (ξ ))∨(x)|2 dt

t

)12

maps L2(Rn) to L2(Rn) with bound at most A√

log2.

6.5.3. ([302]) Let A,a,b > 0 with a+b > 1. Use the idea of Lemma 6.5.2 to show

that if m(ξ ) satisfies |m(ξ )| ≤ A(1+ |ξ |)−a and |∇m(ξ )| ≤ A(1+ |ξ |)−b for all

ξ ∈ Rn, then the maximal operator

Mm( f )(x) = supt>0

∣∣( f (ξ )m(tξ ))∨(x)

∣∣

is bounded from L2(Rn) to itself.[Hint: Use that

Mm ≤∞

∑j=0

Mm, j ,

where Mm, j corresponds to the multiplier ϕ jm; here ϕ j is as in (6.5.8). Show that

∥∥Mm, j( f )∥∥

L2 ≤C∥∥ϕ jm

∥∥ 12

L∞

∥∥ϕ jm∥∥ 1

2

L∞

∥∥ f∥∥

L2≤C 2 j1−(a+b)

2

∥∥ f∥∥

L2 ,

where m(ξ ) = ξ ·∇m(ξ ).]

6.5.4. Let A,c > 0, a > 1/2, 0 < b < n. Follow the idea of the proof of Theorem

6.5.1 to obtain the following more general result: If dµ is a finite Borel measure

supported in the closed unit ball that satisfies |dµ(ξ )| ≤ A(1+ |ξ |)−a for all ξ ∈Rn

and dµ(B(y,R))≤ cRb for all R > 0, then the maximal operator

f → supt>0

∣∣∣∫

Rnf (x− ty)dµ(y)

∣∣∣

maps Lp(Rn) to itself when p > 2n−2b+2a−1n−b+2a−1

.[Hint: Using the notation of the preceding exercise, show that ‖Mm, j( f )‖L2 ≤

C 2 j( 12−a)‖ f‖L2 and that ‖Mm, j( f )‖L1,∞ ≤ C 2 j(n−b)‖ f‖L1 for all j ∈ Z+, where C

is a constant depending on the given parameters.]

6.5.5. Show that Theorem 6.5.1 is false when n = 1, that is, show that the maximal

operator

M1( f )(x) = supt>0

| f (x+ t)+ f (x− t)|2

is unbounded on Lp(R) for all p < ∞.

6.5.6. Show that when n ≥ 2 and p ≤ nn−1

there exists an Lp(Rn) function f such

that M ( f )(x) = ∞ for all x ∈ Rn. Hence Theorem 6.5.1 is false is this case.[Hint: Choose a compactly supported and radial function equal to |y|1−n(− log |y|)−1

when |y| ≤ 1/2.]

6.6 Wavelets and Sampling

In this section we construct orthonormal bases of L2(R) generated by translations

and dilations of a single function. An example of such base is given by the Haar

functions we encountered in Section 6.4. The Haar functions are generated by in-

teger translations and dyadic dilations of the single function χ[0, 12 )− χ[ 1

2 ,1). This

function is not smooth, and the main question addressed in this section is whether

there exist smooth analogues of the Haar functions.

Definition 6.6.1. A square integrable function ϕ on Rn is called a wavelet if the

family of functions

ϕν ,k(x) = 2νn2 ϕ(2νx− k) ,

where ν ranges over Z and k over Zn, is an orthonormal basis of L2(Rn). This

means that the functions ϕν ,k are mutually orthogonal and span L2(Rn), and ϕ is

normalized to have L2 norm equal to 1. Note that the Fourier transform of ϕν ,k is

given by

ϕν ,k(ξ ) = 2−νn2 ϕ(2−νξ )e−2πi2−νξ ·k . (6.6.1)

6.6 Wavelets and Sampling 483

Rephrasing the question posed earlier, the main issue addressed in this section is

whether smooth wavelets actually exist. Before we embark on this topic, we recall

that we have already encountered examples of nonsmooth wavelets.

Example 6.6.2. (The Haar wavelet) Recall the family of functions

hI(x) = |I|−12 (χIL − χIR) ,

where I ranges over D (the set of all dyadic intervals) and IL is the left part of I and

IR is the right part of I. Note that if I = [2−νk,2−ν(k+1)), then

hI(x) = 2ν2 ϕ(2νx− k) ,

where

ϕ(x) = χ[0, 12 )− χ[ 1

2 ,1). (6.6.2)

The single function ϕ in (6.6.2) therefore generates the Haar basis by taking trans-

lations and dilations. Moreover, we observed in Section 6.4 that the family hII is

orthonormal. Moreover, in Theorem 6.4.6 we obtained the representation

f = ∑I∈D

⟨f ,hI

⟩hI in L2 ,

which proves the completeness of the system hII∈D in L2(R).

6.6.1 Some Preliminary Facts

Before we look at more examples, we make some observations. We begin with the

following useful fact.

Proposition 6.6.3. Let g ∈ L1(Rn). Then

g(m) = 0 for all m ∈ Zn \0

if and only if

∑k∈Zn

g(x+ k) =∫

Rng(t)dt

for almost all x ∈ Tn.

Proof. We define the periodic function

G(x) = ∑k∈Zn

g(x+ k) ,


which is easily shown to be in L1(Tn). Moreover, we have

G(m) = g(m)

for all m ∈ Zn, where G(m) denotes the mth Fourier coefficient of G and g(m) de-

notes the Fourier transform of g at ξ = m. If g(m) = 0 for all m ∈ Zn \ 0, then

all the Fourier coefficients of G (except for m = 0) vanish, which means that the

sequence G(m)m∈Zn lies in ℓ1(Zn) and hence Fourier inversion applies. We con-

clude that for almost all x ∈ Tn we have

G(x) = ∑m∈Zn

G(m)e2πim·x = G(0) = g(0) =∫

Rng(t)dt .

Conversely, if G is a constant, then G(m) = 0 for all m ∈ Zn \0, and so the same

holds for g.

A consequence of the preceding proposition is the following.

Proposition 6.6.4. Let ϕ ∈ L2(Rn). Then the sequence

ϕ(x− k)k∈Zn (6.6.3)

forms an orthonormal set in L2(Rn) if and only if

∑k∈Zn

|ϕ(ξ + k)|2 = 1 (6.6.4)

for almost all ξ ∈ Rn.

Proof. Observe that either (6.6.4) or the hypothesis that the sequence in (6.6.3) is

orthonormal implies that ‖ϕ‖L2 = 1. Also the orthonormality condition

∫

Rnϕ(x− j)ϕ(x− k)dx =

1 when j = k,

0 when j = k,

is equivalent to

∫

Rne−2πik·ξ ϕ(ξ )e−2πi j·ξ ϕ(ξ )dξ = (|ϕ|2)(k− j) =

1 when j = k,

0 when j = k,

in view of Parseval’s identity. Proposition 6.6.3 with g(ξ ) = |ϕ(ξ )|2 gives that the

latter is equivalent to

∑k∈Zn

|ϕ(ξ + k)|2 =∫

Rn|ϕ(t)|2 dt = 1

for almost all ξ ∈ Rn.


Corollary 6.6.5. Let ϕ ∈ L2(Rn) and suppose that the sequence

ϕ(x− k)k∈Zn (6.6.5)

forms an orthonormal set in L2(Rn). Then the measure of the support of ϕ is at least

1, that is,

|supp ϕ| ≥ 1 . (6.6.6)

Moreover, if |supp ϕ|= 1, then |ϕ(ξ )|= 1 for almost all ξ ∈ supp ϕ .

Proof. It follows from (6.6.4) that |ϕ| ≤ 1 for almost all ξ ∈ Rn and thus

|supp ϕ| ≥∫

Rn|ϕ(ξ )|2 dξ =

∫

[0,1)n∑

k∈Zn

|ϕ(ξ + k)|2 dξ =∫

[0,1)n1dξ = 1 .

If equality holds in (6.6.6), then equality holds in the preceding inequality, and since

|ϕ| ≤ 1 a.e., it follows that |ϕ(ξ )|= 1 for almost all ξ in supp ϕ .

6.6.2 Construction of a Nonsmooth Wavelet

Having established these preliminary facts, we now start searching for examples of

wavelets. It follows from Corollary 6.6.5 that the support of the Fourier transform of

a wavelet must have measure at least 1. It is reasonable to ask whether this support

can have measure exactly 1. Example 6.6.6 indicates that this can indeed happen. As

dictated by the same corollary, the Fourier transform of such a wavelet must satisfy

|ϕ(ξ )| = 1 for almost all ξ ∈ supp ϕ , so it is natural to look for a wavelet ϕ such

that ϕ = χA for some set A. We can start by asking whether the function

ϕ = χ[− 12 ,

12 ]

on R is an appropriate Fourier transform of a wavelet, but a moment’s thought shows

that the functions ϕµ ,0 and ϕν ,0 cannot be orthogonal to each other when µ = 0.

The problem here is that the Fourier transforms of the functions ϕν ,k cluster near

the origin and do not allow for the needed orthogonality. We can fix this problem

by considering a function whose Fourier transform vanishes near the origin. Among

such functions, a natural candidate is

χ[−1,− 12 )+ χ[ 1

2 ,1), (6.6.7)

which is indeed the Fourier transform of a wavelet.

Example 6.6.6. Let A = [−1,− 12)⋃[ 1

2,1) and define a function ϕ on R by setting

ϕ = χA .


Then we assert that the family of functions

ϕν ,k(x)k∈Z,ν∈Z = 2ν/2ϕ(2νx− k)k∈Z,ν∈Z

is an orthonormal basis of L2(R) (i.e., the function ϕ is a wavelet). This is an exam-

ple of a wavelet with minimally supported frequency.

To verify this assertion, first note that ϕ0,kk∈Z is an orthonormal set, since

(6.6.4) is easily seen to hold. Dilating by 2ν , it follows that ϕν ,kk∈Z is also an

orthonormal set for every fixed ν ∈ Z. Next, observe that if µ = ν , then

supp ϕν ,k ∩ supp ϕµ ,l = /0 . (6.6.8)

This implies that the family 2ν/2ϕ(2νx− k)k∈Z,ν∈Z is also orthonormal.

Next, we observe that the completeness of ϕν ,kν ,k∈Z is equivalent to that of

ϕν ,k(ξ )ν ,k∈Z = 2−ν/2e−2πikξ2−ν χ2νA(ξ )ν ,k∈Z. Let f ∈ L2(R), fix any ν ∈ Z,

and define

h(ξ ) = 2ν/2 f (2νξ ).

Suppose that for all k ∈ Z,

0 = 〈 f , ϕν ,k〉=∫

2νAf (ξ )2−ν/2e−2πikξ2−νdξ

=∫

A2ν/2 f (2νξ )e−2πikξdξ

= 〈χAh,e−2πikξ 〉 .

Exercise 6.6.1(a) shows e−2πikξk∈Z is an orthonormal basis of L2(A), and there-

fore χAh = 0 almost everywhere. From the definition of h it follows that χ2νA f = 0

almost everywhere. Now suppose for all ν ,k ∈ Z

0 = 〈 f , ϕν ,k〉.

Then χ2νA f = 0 almost everywhere for all ν ∈ Z. Since ∪ν∈Z2νA = R \ 0, it

follows that f = 0 almost everywhere. We conclude ϕν ,kν ,k∈Z is complete.

6.6.3 Construction of a Smooth Wavelet

The wavelet basis of L2(Rn) constructed in Example 6.6.6 is forced to have slow

decay at infinity, since the Fourier transforms of the elements of the basis are non-

smooth. Smoothing out the function ϕ but still expecting ϕ to be wavelet is a bit

tricky, since property (6.6.8) may be violated when µ = ν , and moreover, (6.6.4)

may be destroyed. These two obstacles are overcome by the careful construction of

the next theorem.


Theorem 6.6.7. There exists a Schwartz function ϕ on the real line that is a wavelet,

that is, the collection of functions ϕν ,kk,ν∈Z with ϕν ,k(x) = 2ν2 ϕ(2νx− k) is an

orthonormal basis of L2(R). Moreover, the function ϕ can be constructed so that its

Fourier transform satisfies

supp ϕ ⊆[− 4

3,− 1

3

]∪[

13, 4

3

]. (6.6.9)

Note that in view of condition (6.6.9), the function ϕ must have vanishing mo-

ments of all orders.

Proof. We start with an odd smooth real-valued functionΘ on the real line such that

Θ(t) = π4

for t ≥ 16

and such that Θ is strictly increasing on the interval[− 1

6, 1

6

].

We set

α(t) = sin(Θ(t)+ π4), β (t) = cos(Θ(t)+ π

4),

and we observe that

α(t)2 +β (t)2 = 1

and that

α(−t) = β (t)

for all real t. Next we introduce the smooth function ω defined via

ω(t) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

β (− t2− 1

2) = α( t

2+ 1

2) when t ∈

[− 4

3,− 2

3

],

α(−t− 12) = β (t + 1

2) when t ∈

[− 2

3,− 1

3

],

α(t− 12) when t ∈

[13, 2

3

],

β ( t2− 1

2) when t ∈

[23, 4

3

],

on the interval[− 4

3,− 1

3

]⋃[13, 4

3

]. Note thatω is an even function. Finally we define

the function ϕ by letting

ϕ(ξ ) = e−πiξω(ξ ) ,

and we note that

ϕ(x) =∫

Rω(ξ )e2πiξ (x− 1

2 )dξ = 2

∫ ∞

0ω(ξ )cos

(2π(x− 1

2)ξ

)dξ .

It follows that the function ϕ is symmetric about the number 12, that is, we have

ϕ(x) = ϕ(1− x)

for all x ∈ R. Note that ϕ is a Schwartz function whose Fourier transform is sup-

ported in the set[− 4

3,− 1

3

]⋃[13, 4

3

].

Having defined ϕ , we proceed by showing that it is a wavelet. In view of identity

(6.6.1) we have that ϕν ,k is supported in the set 132ν ≤ |ξ | ≤ 4

32ν , while ϕµ , j is

supported in the set 132µ ≤ |ξ | ≤ 4

32µ . The intersection of these sets has measure

zero when |µ − ν | ≥ 2, which implies that such wavelets are orthogonal to each

other. Therefore, it suffices to verify orthogonality between adjacent scales (i.e.,

when ν = µ and ν = µ+1).


We begin with the case ν = µ , which, by a simple dilation, is reduced to the case

ν = µ = 0. Thus to obtain the orthogonality of the functions ϕ0,k(x) = ϕ(x−k) and

ϕ0, j(x) = ϕ(x− j), in view of Proposition 6.6.4, it suffices to show that

∑k∈Z

|ϕ(ξ + k)|2 = 1 . (6.6.10)

Since the sum in (6.6.10) is 1-periodic, we check that is equal to 1 only for ξ in[13, 4

3

]. First for ξ ∈

[13, 2

3

], the sum in (6.6.10) is equal to

|ϕ(ξ )|2 + |ϕ(ξ −1)|2 = ω(ξ )2 +ω(ξ −1)2

= α(ξ − 12)2 +β ((ξ −1)+ 1

2)2

= 1

from the definition of ω . A similar argument also holds for ξ ∈[

23, 4

3

], and this

completes the proof of (6.6.10). As a consequence of this identity we also obtain

that the functions ϕ0,k have L2 norm equal to 1, and thus so have the functions ϕν ,k,

via a change of variables.

Next we prove the orthogonality of the functions ϕν ,k and ϕν+1, j for general

ν ,k, j ∈ Z. We begin by observing the validity of the following identity:

ϕ(ξ )ϕ( ξ2) =

e−πiξ/2β ( ξ

2− 1

2)α( ξ

2− 1

2) when 2

3≤ ξ ≤ 4

3,

e−πiξ/2α( ξ2+ 1

2)β ( ξ

2+ 1

2) when − 4

3≤ ξ ≤− 2

3.

(6.6.11)

Indeed, from the definition of ϕ , it follows that

ϕ(ξ )ϕ( ξ2) = e−πiξ/2ω(ξ )ω( ξ

2) .

This function is supported in

ξ ∈ R : 13≤ |ξ | ≤ 4

3∩ξ ∈ R : 2

3≤ |ξ | ≤ 8

3= ξ ∈ R : 2

3≤ |ξ | ≤ 4

3 ,

and on this set it is equal to

e−πiξ/2

β ( ξ

2− 1

2)α( ξ

2− 1

2) when 2

3≤ ξ ≤ 4

3,

α( ξ2+ 1

2)β ( ξ

2+ 1

2) when − 4

3≤ ξ ≤− 2

3,

by the definition of ω . This establishes (6.6.11).

We now turn to the orthogonality of the functions ϕν ,k and ϕν+1, j for general

ν ,k, j ∈ Z. Using (6.6.1) and (6.6.11) we have

⟨ϕν ,k |ϕν+1, j

⟩=

⟨ϕν ,k | ϕν+1, j

⟩

=

∫

R2−

ν2 ϕ(2−νξ )e−2πi

ξ k

2ν 2−ν+1

2 ϕ(2−(ν+1)ξ )e−2πi

ξ j

2ν+1 dξ


=1√2

∫

Rϕ(ξ )ϕ( ξ

2)e−2πiξ (k− j

2 ) dξ

=1√2

∫ − 23

− 43

α( ξ2+ 1

2)β ( ξ

2+ 1


2+14 ) dξ

+1√2

∫ 43

23

α( ξ2− 1

2)β ( ξ

2− 1


2+14 ) dξ

= 0 ,

where the last identity follows from the change of variables ξ = ξ ′−2 in the second-

to-last integral, which transforms its range of integration to[

23, 4

3

]and its integrand

to the negative of that of the last displayed integral.

Our final task is to show that the orthonormal system ϕν ,kν ,k∈Z is complete.

We show this by proving that whenever a square-integrable function f satisfies

⟨f |ϕν ,k

⟩= 0 (6.6.12)

for all ν ,k ∈ Z, then f must be zero. Suppose that (6.6.12) holds. Plancherel’s iden-

tity yields ∫

Rf (ξ )2−

ν2 ϕ(2−νξ )e−2πi2−νξk dξ = 0

for all ν ,k and thus

∫

Rf (2νξ )ϕ(ξ )e2πiξk dξ =

(f (2ν(·)) ϕ

)(−k) = 0 (6.6.13)

for all ν ,k ∈ Z. It follows from Proposition 6.6.3 and (6.6.13) (with k = 0) that

∑k∈Z

f (2ν(ξ + k))ϕ(ξ + k) =

∫

Rf (2νξ ) ϕ(ξ )dξ =

(f (2ν(·)) ϕ

)(0) = 0

for all ν ∈ Z.

Next, we show that the identity

∑k∈Z

f (2ν(ξ + k))ϕ(ξ + k) = 0 (6.6.14)

for all ν ∈ Z implies that f is identically equal to zero. Suppose that 13≤ ξ ≤ 2

3. In

this case the support properties of ϕ imply that the only terms in the sum in (6.6.14)

that do not vanish are k = 0 and k =−1. Thus for 13≤ ξ ≤ 2

3the identity in (6.6.14)

reduces to

0 = f (2ν(ξ −1))ϕ(ξ −1)+ f (2νξ )ϕ(ξ )

= f (2ν(ξ −1))eπi(ξ−1)β ((ξ −1)+ 12)+ f (2νξ )eπiξα(ξ − 1

2) ;


hence

− f (2ν(ξ −1))β (ξ − 12)+ f (2νξ )α(ξ − 1

2) = 0, 1

3≤ ξ ≤ 2

3. (6.6.15)

Next we observe that when 23≤ ξ ≤ 4

3, only the terms with k = 0 and k =−2 survive

in the identity in (6.6.14). This is because when k =−1, ξ+k = ξ−1∈[− 1

3, 1

3

]and

this interval has null intersection with the support of ϕ . Therefore, (6.6.14) reduces

to

0 = f (2ν(ξ −2))ϕ(ξ −2)+ f (2νξ )ϕ(ξ )

= f (2ν(ξ −2))eπi(ξ−2)α( ξ−22

+ 12)+ f (2νξ )eπiξβ ( ξ

2− 1

2) ;

hence

f (2ν(ξ −2))α( ξ2− 1

2)+ f (2νξ )β ( ξ

2− 1

2) = 0, 2

3≤ ξ ≤ 4

3. (6.6.16)

Replacing first ν by ν−1 and thenξ2

by ξ in (6.6.16), we obtain

f (2ν(ξ −1))α(ξ − 12)+ f (2νξ )β (ξ − 1

2) = 0, 1

3≤ ξ ≤ 2

3. (6.6.17)

Now consider the 2× 2 system of equations given by (6.6.15) and (6.6.17) with

unknown f (2ν(ξ −1)) and f (2νξ ). The determinant of the system is

det

(−β (ξ −1/2) α(ξ −1/2)α(ξ −1/2) β (ξ −1/2)

)=−1 = 0 .

Therefore, the system has the unique solution

f (2ν(ξ −1)) = f (2νξ ) = 0 ,

which is valid for all ν ∈Z and all ξ ∈ [ 13, 2

3]. We conclude that f (ξ )= 0 for all ξ ∈R

and thus f = 0. This proves the completeness of the system ϕν ,k. We conclude that

the function ϕ is a wavelet.

6.6.4 Sampling

Next we discuss how one can recover a band-limited function by its values at a

countable number of points.

Definition 6.6.8. An integrable function on Rn is called band limited if its Fourier

transform has compact support.

For every band-limited function there is a B > 0 such that its Fourier transform

is supported in the cube [−B,B]n. In such a case we say that the function is band

limited on the cube [−B,B]n.

It is an interesting observation that such functions are completely determined by

their values at the points x = k/2B, where k ∈ Zn. We have the following result.

Theorem 6.6.9. (a) Let f in L1(Rn) be band limited on the cube [−B,B]n. Then f

can be sampled by its values at the points x = k/2B, where k ∈ Zn. In particular, we

have

f (x1, . . . ,xn) = ∑k∈Zn

f( k

2B

) n

∏j=1

sin(2πBx j−πk j)

2πBx j−πk j

(6.6.18)


(b) Suppose that f is band-limited on the cube [−B′,B′]n where 0 < B′ < B. Then f

can be sampled by its values at the points x = k/2B, k ∈ Zn as follows

f (x1, . . . ,xn) = ∑k∈Zn

f( k

2B

)Φ(x− k) , (6.6.19)

for some Schwartz function Φ that depends on B,B′.

Proof. Since the function f is supported in [−B,B]n, we use Exercise 6.6.2 to obtain

f (ξ ) =1

(2B)n ∑k∈Zn

f( k

2B

)e2πi k

2B ·ξ

=1

(2B)n ∑k∈Zn

f(− k

2B

)e2πi k

2B ·ξ .

Inserting this identity in the inversion formula

f (x) =∫

[−B,B]nf (ξ )e2πix·ξ dξ ,

which holds for almost all x∈Rn since f is continuous and therefore integrable over

[−B,B]n, we obtain

f (x) =∫

[−B,B]n

1

(2B)n ∑k∈Zn

f(− k

2B

)e2πi k

2B ·ξ e2πix·ξ dξ

= ∑k∈Zn

f(− k

2B

) 1

(2B)n

∫

[−B,B]ne2πi( k

2B+x)·ξ dξ (6.6.20)

= ∑k∈Zn

f(− k

2B

) n

∏j=1

sin(2πBx j +πk j)

2πBx j +πk j

. (6.6.21)

This is exactly (6.6.18) when we change k to −k and thus part (a) is proved. For part

(b) we argue similarly, except that we replace χ[−B,B]n by Φ , where Φ is smooth,

equal to 1 on [−B′,B′]n and vanishes outside [−B,B]n. Then we can insert the func-

tion Φ(ξ ) in (6.6.20) and instead of (6.6.21) we obtain the expression on the right

in (6.6.19).

Remark 6.6.10. Identity (6.6.18) holds for any B′′ > B. In particular, we have

∑k∈Zn

f( k

2B

) n

∏j=1


2πBx j−πk j

= ∑k∈Zn

f( k

2B′′

) n

∏j=1

sin(2πB′′x j−πk j)

2πB′′x j−πk j

for all x ∈ Rn whenever f is band-limited in [−B,B]n. In particular, band-limited

functions in [−B,B]n can be sampled by their values at the points k/2B′′ for any

B′′ ≥ B.

However, band-limited functions in [−B,B]n cannot be sampled by the points

k/2B′ for any B′ < B, as the following example indicates.

Example 6.6.11. For 0 < B′ < B, let f (x) = g(x)sin(2πB′x), where g is supported

in the interval [−(B−B′),B−B′]. Then f is band limited in [−B,B], but it cannot

be sampled by its values at the points k/2B′, since it vanishes at these points and f

is not identically zero if g is not the zero function.

Next, we give a couple of results that relate the Lp norm of a given function with

the ℓp norm (or quasi-norm) of its sampled values.

Theorem 6.6.12. Let f be a tempered1 function whose Fourier transform is sup-

ported in the closed ball B(0, t) for some 0 < t < ∞. Assume that f lies in Lp(Rn)for some 0 < p≤ ∞. Then there is a constant C(n, p) such that

∥∥ f (k)k∈Zn

∥∥ℓp(Zn)

≤C(n, p) t (1+ t2np )

∥∥ f∥∥

Lp(Rn).

Proof. The proof is based on the following fact, whose proof can be found in [131]

(Lemma 2.2.3). Let 0 < r < ∞. Then there exists a constant C2 =C2(n,r) such that

for all t > 0 and for all C 1 functions u on Rn whose distributional Fourier transform

is supported in the ball |ξ | ≤ t we have

supz∈Rn

1

t

|∇u(x− z)|(1+ t|z|) n

r

≤C2 M(|u|r)(x) 1r , (6.6.22)

where M denotes the Hardy–Littlewood maximal operator.

Notice that f is a C ∞ function since its Fourier transform is compactly supported.

Assuming (6.6.22), for each k ∈ Zn and x ∈ [0,1]n we use the mean value theorem

to obtain

| f (k)| ≤ | f (x+ k)|+√

n supz∈[0,1]n

|∇ f (z+ k)|

≤ | f (x+ k)|+√

n supz∈B(x+k,

√n)

|∇ f (z)| .

We raise this inequality to the power p, we integrate over the cube [0,1]n, we sum

over k∈Zn, and then we take the 1/p power. Let cp =max(1,21/p−1) and c(n,r, t) =

1 A function is called tempered if there are constants C,M such that | f (x)| ≤ C (1+ |x|)M for all

x ∈ Rn. Tempered functions are tempered distributions.

√nt(1+ t

√n)n/r. The sum over k and the integral over [0,1]n yield an integral over

Rn and thus we obtain

[∑

k∈Zn

| f (k)|p] 1

p ≤[∫

Rn| f (x)+

√n sup

z∈B(x,√

n)

|∇ f (z)|p dx

] 1p

≤ cp

[∥∥ f∥∥

Lp +√

n

(∫

Rnsup

z∈B(0,√

n)

|∇ f (x− z)|p dx

) 1p]

≤ cp

[∥∥ f∥∥

Lp + c(n,r, t)

(∫

Rn

sup

z∈B(0,√

n)

|∇ f (x− z)|t(1+ t|z|) n

r

p

dx

) 1p]

≤ cp

[∥∥ f∥∥

Lp + c(n,r, t)

(∫

Rn

supz∈Rn

|∇ f (x− z)|t(1+ t|z|) n

r

p

dx

) 1p]

≤ cp

[∥∥ f∥∥

Lp + c(n,r, t)C2

(∫

Rn[M(| f |r)(x)]

pr dx

) 1p],

where the last step uses (6.6.22). We now select r = p/2 if p < ∞ and r to be

any number if p = ∞. The required inequality follows from the boundedness of the

Hardy-Littlewood maximal operator on L2 if p < ∞ or on L∞ if p = ∞.

The next theorem could be considered a partial converse of Theorem 6.6.13

Theorem 6.6.13. Suppose that an integrable function f has Fourier transform sup-

ported in the cube [−( 12− ε), 1

2− ε ]n for some 0 < ε < 1/2. Furthermore, suppose

that the sequence of coefficients f (k)k∈Zn lies in ℓp(Zn) for some 0 < p≤∞. Then

f lies in Lp(Rn) and the following estimate is valid

∥∥ f∥∥

Lp(Rn)≤Cn,p,ε

∥∥ f (k)k

∥∥ℓp(Zn)

. (6.6.23)

Proof. We fix a smooth function Φ supported in [− 12, 1

2]n and equal to 1 on the

smaller cube [−( 12− ε), 1

2− ε ]n. Then we may write f = f ∗Φ , since Φ is equal to

one on the support of f . Writing f in terms of its Fourier series we have

f (ξ ) = ∑k∈Zn

f (k)e2πik·ξ χ[− 1

2 ,12 ]

n = ∑k∈Zn

f (−k)e2πik·ξ χ[− 12 ,

12 ]

n (6.6.24)

Since f is integrable, f is continuous and thus integrable over [− 12, 1

2]n. By Fourier

inversion we have

f (x) =∫

[− 12 ,

12 ]

nf (ξ )e2πix·ξdξ =

∫

[− 12 ,

12 ]

nf (ξ )Φ(ξ )e2πix·ξdξ (6.6.25)

for almost all x ∈ Rn. Inserting (6.6.25) in (6.6.24) we obtain

f (x) =∫

[− 12 ,

12 ]

n∑

k∈Zn

f (−k)e2πik·ξ e2πix·ξ Φ(ξ )dξ

= ∑k∈Zn

f (k)∫

[− 12 ,

12 ]

ne−2πik·ξ e2πix·ξ Φ(ξ )dξ

= ∑k∈Zn

f (k)Φ(x− k) .

This identity combined with the rapid decay of Φ yields (6.6.23) as follows. For

0 < p≤ 1 we have

∥∥ f∥∥p

Lp ≤∫

Rn∑

k∈Zn

| f (k)|p|Φ(x− k)|p =∥∥ f (k)k‖p

ℓp(Zn)‖Φ‖p

Lp

while for 1 < p≤ ∞, setting Q = [− 12, 1

2]n we write:

∥∥ f∥∥

Lp(Rn)≤

[∑

l∈Zn

∫

l+Q

(∑

k∈Zn

| f (k)||Φ(x− k)|)p

dx

] 1p

≤ Cn,N

[∑

l∈Zn

∫

l+Q

(∑

k∈Zn

| f (k)| 1

(2√

n+ |x− k|)N

)p

dx

] 1p

≤ C′n,N

[∑

l∈Zn

∫

l+Q

(∑

k∈Zn

| f (k)| 1

(√

n+ |l− k|)N

)p

dx

] 1p

≤ C′n,N

[∑

l∈Zn

(∑

k∈Zn

| f (k)| 1

(√

n+ |l− k|)N

)p] 1

p

.

The preceding expression can be viewed as the ℓp norm of the discrete convolution

of the sequences f (k)k and 1(√

n+|k|)N and thus it is bounded by a constant multiple

of∥∥ f (k)k

∥∥ℓp(Zn)

, since the sequence 1(√

n+|k|)N is in ℓ1(Zn) if N is large enough.

This completes the proof.

Exercise 6.6.6 gives examples of functions for which Theorem 6.6.13 fails

if ε = 0.

Exercises

6.6.1. (a) Let A = [−1,− 12)⋃[ 1

2,1). Show that the family e2πimxm∈Z is an

orthonormal basis of L2(A).(b) Obtain the same conclusion for the family e2πim·xm∈Zn in L2(An).[Hint: To show completeness, given f ∈ L2(A), define h on [0,1] by setting h(x) =

f (x−1) for x ∈ [0, 12) and h(x) = f (x) for x ∈ [ 1

2,1). Observe that h(m) = f (m) for

all m ∈ Z and expand h in Fourier series.]

6.6.2. Let g be an integrable function on Rn.

(a) Suppose that g is supported in [−b,b]n for some b > 0 and that the sequence

g(k/2b)k∈Zn lies in ℓ2(Zn). Show that

g(x) = (2b)−n ∑k∈Zn

g( k2b)e2πi k

2b·xχ[−b,b]n ,

where the series converges in L2(Rn) and deduce that g is in L2(Rn).(b) Suppose that g is supported in [0,b]n for some b > 0 and that the sequence

g(k/b)k∈Zn lies in ℓ2(Zn). Show that

g(x) = b−n ∑k∈Zn

g( kb)e2πi k

b·xχ[0,b]n ,

where the series converges in L2(Rn) and deduce that g is in L2(Rn).(c) When n = 1, obtain the same as the conclusion in part (b) for x ∈ [−b,− b

2)⋃

[ b2,b), provided g is supported in this set.[

Hint: Part (c): Use the result in Exercise 6.6.1.]

6.6.3. Show that the sequence of functions

Hk(x1, . . . ,xn) = (2B)n2

n

∏j=1

sin(π(2Bx j− k j)

)

π(2Bx j− k j), k ∈ Zn ,

is orthonormal in L2(Rn).[Hint: Interpret the functions Hk as the Fourier transforms of known functions.

]

6.6.4. Prove the following spherical multidimensional version of Theorem 6.6.9.

Suppose that f is supported in the ball |ξ | ≤ R. Show that

f (x) = ∑k∈Zn

f(− k

2R

) 1

2n

J n2(2π|Rx+ k

2|)

|Rx+ k2| n

2

,

where Ja is the Bessel function of order a.

6.6.5. Let akk∈Zn be in ℓp for some 1 < p < ∞. Show that the partial sums

∑k∈Zn

|k|≤N

ak

n

∏j=1


2πBx j−πk j

converge in S ′(Rn) as N → ∞ to an Lp function on Rn whose Fourier transform

is supported in [−B,B]n. Here k = (k1, . . . ,kn). Moreover, the Lp norm of A is con-

trolled by a constant multiple of the ℓp norm of akk.

6.6.6. Consider the function ∏nj=1 sin(πx j)/(πx j) on Rn to show that Theorem

6.6.13 fails when ε = 0 and p ≤ 1. When 1 < p ≤ ∞ consider the function x1 +

∏nj=1 sin(πx j)/(πx j).

6.6.7. (a) Let ψ(x) be a nonzero continuous integrable function on R that satisfies∫Rψ(x)dx = 0 and

Cψ =∫ +∞

−∞

|ψ(t)|2|t| dt < ∞ .

Define the wavelet transform of f in L2(R) by setting

W ( f ;a,b) =1√|a|

∫ +∞

−∞f (x)ψ

(x−b

a

)dx

when a = 0 and W ( f ;0,b) = 0. Show that for any f ∈ L2(R) the following inversion

formula holds:

f (x) =1

Cψ

∫ +∞

−∞

∫ +∞

−∞

1

|a| 12

ψ(x−b

a

)W ( f ;a,b)db

da

a2.

(b) State and prove an analogous wavelet transform inversion property on Rn.[Hint: Apply Theorem 2.2.14 (5) in the b-integral and use Fourier inversion.

]

6.6.8. (P. Casazza) On Rn let e j be the vector whose coordinates are zero every-

where except for the jth entry, which is 1. Set q j = e j− 1n ∑

nk=1 ek for 1≤ j ≤ n and

also qn+1 =1√n ∑

nk=1 ek. Prove that

n+1

∑j=1

|q j · x|2 = |x|2

for all x ∈ Rn. This provides an example of a tight frame on Rn.

HISTORICAL NOTES

An early account of square functions in the context of Fourier series appears in the work of

Kolmogorov [196], who proved the almost everywhere convergence of lacunary partial sums of

Fourier series of periodic square-integrable functions. This result was systematically studied and

extended to Lp functions, 1 < p < ∞, by Littlewood and Paley [227], [228], [229] using complex-

analysis techniques. The real-variable treatment of the Littlewood and Paley theorem was pioneered

by Stein [334] and allowed the higher-dimensional extension of the theory. The use of vector-valued

inequalities in the proof of Theorem 6.1.2 is contained in Benedek, Calderon, and Panzone [22]. A

Littlewood–Paley theorem for lacunary sectors in R2 was obtained by Nagel, Stein, and Wainger

[264].

An interesting Littlewood–Paley estimate holds for 2≤ p < ∞: There exists a constant Cp such

that for all families of disjoint open intervals I j in R the estimate ‖(∑ j |( f χI j)∨|2) 1

2 ‖Lp ≤Cp‖ f‖Lp

holds for all functions f ∈ Lp(R). This was proved by Rubio de Francia [301], but the special case

in which I j = ( j, j+1) was previously obtained by Carleson [55]. An alternative proof of Rubio de

Francia’s theorem was obtained by Bourgain [34]. A higher-dimensional analogue of this estimate

for arbitrary disjoint open rectangles in Rn with sides parallel to the axes was obtained by Journe

[181]. Easier proofs of the higher-dimensional result were subsequently obtained by Sjolin [326],

Soria [329], and Sato [311].

Part (a) of Theorem 6.2.7 is due to Mihlin [254] and the generalization in part (b) to Hormander

[159]. Theorem 6.2.2 can be found in Marcinkiewicz’s article [241] in the context of one-

dimensional Fourier series. Calderon and Torchinsky [45] have improved Theorem 6.2.7 in the

following way: if for a suitable smooth bump η supported in an annulus the functions m(2kξ )η(ξ )lie in the Sobolev space Lr

γ uniformly in k ∈ Z, where γ > n( 1p− 1

2), 1 < p < 2, 1

r= 1

p− 1

2, then m

lies in Mp(Rn). The power 6 in estimate (6.2.3) that appears in the statement of Theorem 6.2.2 is

not optimal. Tao and Wright [357] proved that in dimension 1, the best power of (p−1)−1 in this

theorem is 32

as p→ 1. An improvement of the Marcinkiewicz multiplier theorem in one dimen-

sion was obtained by Coifman, Rubio de Francia, and Semmes [69]. Weighted norm estimates for

Hormander–Mihlin multipliers were obtained by Kurtz and Wheeden [209] and for Marcinkiwiecz

multipliers by Kurtz [208]. Heo, Nazarov, and Seeger [150] have obtained a very elegant charac-

terization of radial Lp multipliers in large dimensions; precisely, they showed that for dimensions

n≥ 4 and 1 0 tn/p‖(m( ·)η(t ·))∨‖Lp <∞. This characteri-

zation builds on and extends a previously obtained simple characterization by Garrigos and Seeger

[124] of radial multipliers on the invariant subspace of radial Lp functions when 1 < p < 2nn+1

.

The method of proof of Theorem 6.3.4 is adapted from Duoandikoetxea and Rubio de Francia

[102]. The method in this article is rather general and can be used to obtain Lp boundedness for a

variety of rough singular integrals. A version of Theorem 6.3.6 was used by Christ [59] to obtain

Lp smoothing estimates for Cantor–Lebesgue measures. When p = q = 2, Theorem 6.3.6 is false

in general, but it is true for all r satisfying | 1r− 1

2|< | 1

p− 1

2| under the additional assumption that

the m j’s are Lipschitz functions uniformly at all scales. This result was independently obtained

by Carbery [52] and Seeger [316]. Miyachi [255] has obtained a complete characterization of the

indices a,b> 0 such that the functions |x|−bei|x|aψ(x) are Lp Fourier multipliers; hereψ is a smooth

function that is equal to 1 near infinity and vanishes near zero.

The probabilistic notions of conditional expectations and martingales have a strong connection

with the Littlewood–Paley theory discussed in this chapter. For the purposes of this exposition we

considered only the case of the sequence of σ -algebras generated by the dyadic cubes of side length

2−k in Rn. The Lp boundedness of the maximal conditional expectation (Doob [97]) is analogous

to the Lp boundedness of the dyadic maximal function; likewise with the corresponding weak type

(1,1) estimate. The Lp boundedness of the dyadic martingale square function was obtained by

Burkholder [39] and is analogous to Theorem 6.1.2. Moreover, the estimate∥∥supk |Ek( f )|

∥∥Lp ≈∥∥S( f )

∥∥Lp , 0 < p <∞, obtained by Burkholder and Gundy [40] and also by Davis [90] is analogous

to the square-function characterization of the Hardy space H p norm. For an exposition on the

different and unifying aspects of Littlewood–Paley theory we refer to Stein [337]. The proof of

Theorem 6.4.8, which quantitatively expresses the almost orthogonality of the Littlewood–Paley

and the dyadic martingale difference operators, is taken from Grafakos and Kalton [133].

The use of quadratic expressions in the study of certain maximal operators has a long history.

We refer to the article of Stein [340] for a historical survey. Theorem 6.5.1 was first proved by Stein

[339]. The proof in the text is taken from an article of Rubio de Francia [302]. Another proof when

n ≥ 3 is due to Cowling and Mauceri [76]. The more difficult case n = 2 was settled by Bour-

gain [36] about 10 years later. Alternative proofs when n = 2 were given by Mockenhaupt, Seeger,

and Sogge [256] as well as Schlag [313]. The boundedness of maximal operators associated to

more general smooth measures on compact surfaces of finite type were investigated by Iosevich

and Sawyer [173]. The powerful machinery of Fourier integral operators was used by Sogge [328]

to obtain the boundedness of spherical maximal operators on compact manifolds without bound-

ary and positive injectivity radius; a simple proof for the boundedness of the spherical maximal

function on the sphere was given by Nguyen [269]. Weighted norm inequalities for the spheri-

cal maximal operator were obtained by Duoandikoetxea and Vega [103]. The discrete spherical

maximal function was studied by Magyar, Stein, and Wainger [237].

Much of the theory of square functions and the ideas associated with them has analogues in the

dyadic setting. A dyadic analogue of the theory discussed here can be obtained. For an introduction

to the area of dyadic harmonic analysis, we refer to Pereyra [276].

The idea of expressing (or reproducing) a signal as a weighted average of translations and

dilations of a single function appeared in early work of Calderon [42]. This idea is in some sense a

forerunner of wavelets. An early example of a wavelet was constructed by Stromberg [352] in his


search for unconditional bases for Hardy spaces. Another example of a wavelet basis was obtained

by Meyer [249]. The construction of an orthonormal wavelet presented in Theorem 6.6.7 is in

Lemarie and Meyer [216]. A compactly supported wavelet was constructed by Daubechies [88].

Mallat [238] introduced the notion of multiresolution analysis, which led to a systematic production

of wavelets. Theorem 6.6.9 is Shannon’s [319] version of Nyquist’s theorem [270] and is referred

to as the Nyquist-Shannon sampling theorem. It is a fundamental result in telecommunications and

signal processing, since it describes how to reconstruct a signal that contains no frequencies higher

than B Hertz in terms of its values at a sequence of points spaced 1/(2B) seconds apart.

The area of wavelets has taken off significantly since its inception, spurred by these early re-

sults. A general theory of wavelets and its use in Fourier analysis was carefully developed in the

two-volume monograph of Meyer [250], [251] and its successor Meyer and Coifman [253]. For

further study and a deeper account of developments on the subject the reader may consult the

books of Daubechies [89], Chui [64], Wickerhauser [374], Kaiser [184], Benedetto and Frazier

[23], Hernandez and Weiss [151], Wojtaszczyk [379], Mallat [239], Meyer [252], Frazier [120],

Grochenig [140], and the references therein. Theorems 6.6.12 and 6.6.13 first appeared in a com-

bined form in the work of Plancherel and Polya [285] for restrictions of entire functions of expo-

nential type on the real line.

Chapter 7

Weighted Inequalities

Weighted inequalities arise naturally in Fourier analysis, but their use is best justified

by the variety of applications in which they appear. For example, the theory of

weights plays an important role in the study of boundary value problems for

Laplace’s equation on Lipschitz domains. Other applications of weighted inequal-

ities include extrapolation theory, vector-valued inequalities, and estimates for cer-

tain classes of nonlinear partial differential equations.

The theory of weighted inequalities is a natural development of the principles and

methods we have acquainted ourselves with in earlier chapters. Although a variety

of ideas related to weighted inequalities appeared almost simultaneously with the

birth of singular integrals, it was only in the 1970s that a better understanding of

the subject was obtained. This was spurred by Muckenhoupt’s characterization of

positive functions w for which the Hardy–Littlewood maximal operator M maps

Lp(Rn,w(x)dx) to itself. This characterization led to the introduction of the class

Ap and the development of weighted inequalities. We pursue exactly this approach

in the next section to motivate the introduction of the Ap classes.

7.1 The Ap Condition

A weight is a nonnegative locally integrable function on Rn that takes values in

(0,∞) almost everywhere. Therefore, weights are allowed to be zero or infinite only

on a set of Lebesgue measure zero. Hence, if w is a weight and 1/w is locally inte-

grable, then 1/w is also a weight.

Given a weight w and a measurable set E, we use the notation

w(E) =∫

Ew(x)dx



499

500 7 Weighted Inequalities

to denote the w-measure of the set E. Since weights are locally integrable functions,

w(E)<∞ for all sets E contained in some ball. The weighted Lp spaces are denoted

by Lp(Rn,w) or simply Lp(w). Recall the uncentered Hardy–Littlewood maximal

operators on Rn over balls

M( f )(x) = supB∋x

AvgB

| f |= supB∋x

1

|B|

∫

B| f (y)|dy ,

and over cubes

Mc( f )(x) = supQ∋x

AvgQ

| f |= supQ∋x

1

|Q|

∫

Q| f (y)|dy ,

where the suprema are taken over all balls B and cubes Q (with sides parallel to the

axes) that contain the given point x. A classical result (Theorem 2.1.6) states that for

all 1 0 such that

∫

RnM( f )(x)p dx≤Cp(n)

p

∫

Rn| f (x)|p dx (7.1.1)

for all functions f ∈ Lp(Rn). We are concerned with the situation in which the mea-

sure dx in (7.1.1) is replaced by w(x)dx for some weight w(x).

7.1.1 Motivation for the Ap Condition

The question we raise is whether there is a characterization of all weights w(x) such

that the strong type (p, p) inequality

∫

RnM( f )(x)p w(x)dx≤Cp

p

∫

Rn| f (x)|p w(x)dx (7.1.2)

is valid for all f ∈ Lp(w).Suppose that (7.1.2) is valid for some weight w and all f ∈ Lp(w) for some

1 < p < ∞. Apply (7.1.2) to the function f χB supported in a ball B and use that

AvgB | f | ≤M( f χB)(x) for all x ∈ B to obtain

w(B)(

AvgB

| f |)p ≤

∫

BM( f χB)

p wdx≤Cpp

∫

B| f |p wdx . (7.1.3)

It follows that

(1

|B|

∫

B| f (t)|dt

)p

≤ Cpp

w(B)

∫

B| f (x)|p w(x)dx (7.1.4)

for all balls B and all functions f . At this point, it is tempting to choose a function

such that the two integrands are equal. We do so by setting f = w−p′/p, which gives

7.1 The Ap Condition 501

f pw = w−p′/p. Under the assumption that infB w > 0 for all balls B, it would follow

from (7.1.4) that

supB balls

(1

|B|

∫

Bw(x)dx

)(1

|B|

∫

Bw(x)−

1p−1 dx

)p−1

≤Cpp . (7.1.5)

If infB w = 0 for some balls B, we take f = (w+ ε)−p′/p to obtain

(1

|B|

∫

Bw(x)dx

)(1

|B|

∫

B(w(x)+ ε)−

p′p dx

)p(1

|B|

∫

B

w(x)dx

(w(x)+ ε)p′

)−1

≤Cpp (7.1.6)

for all ε > 0. Replacing w(x)dx by (w(x)+ ε)dx in the last integral in (7.1.6) we

obtain a smaller expression, which is also bounded by Cpp . Since −p′/p =−p′+1,

(7.1.6) implies that

(1

|B|

∫

Bw(x)dx

)(1

|B|

∫

B(w(x)+ ε)−

p′p dx

)p−1

≤Cpp , (7.1.7)

from which we can still deduce (7.1.5) via the Lebesgue monotone convergence the-

orem by letting ε → 0. We have now obtained that every weight w that satisfies

(7.1.2) must also satisfy the rather strange-looking condition (7.1.5), which we refer

to in the sequel as the Ap condition. It is a remarkable fact, to be proved in this chap-

ter, that the implication obtained can be reversed, that is, (7.1.2) is a consequence

of (7.1.5). This is the first significant achievement of the theory of weights [i.e., a

characterization of all functions w for which (7.1.2) holds]. This characterization is

based on some deep principles discussed in the next section and provides a solid

motivation for the introduction and careful examination of condition (7.1.5).

Before we study the converse statements, we consider the case p = 1. Assume

that for some weight w the weak type (1,1) inequality

w(x ∈ Rn : M( f )(x)> α

)≤ C1

α

∫

Rn| f (x)|w(x)dx (7.1.8)

holds for all functions f ∈ L1(Rn). Since M( f )(x)≥AvgB | f | for all x∈ B, it follows

from (7.1.8) that for all α < AvgB | f | we have

w(B)≤ w(x ∈ Rn : M( f )(x)> α

)≤ C1

α

∫

Rn| f (x)|w(x)dx . (7.1.9)

Taking f χB instead of f in (7.1.9), we deduce that

AvgB

| f |= 1

|B|

∫

B| f (t)|dt ≤ C1

w(B)

∫

B| f (x)|w(x)dx (7.1.10)

for all functions f and balls B. Taking f = χS, we obtain

|S||B| ≤C1

w(S)

w(B), (7.1.11)

where S is any measurable subset of the ball B.

Recall that the essential infimum of a function w over a set E is defined as

ess.infE

(w) = inf

b > 0 : |x ∈ E : w(x)< b|> 0.

Then for every a > ess.infB(w) there exists a subset Sa of B with positive measure

such that w(x)< a for all x ∈ Sa. Applying (7.1.11) to the set Sa, we obtain

1

|B|

∫

Bw(t)dt ≤ C1

|Sa|

∫

Sa

w(t)dt ≤C1a , (7.1.12)

which implies

1

|B|

∫

Bw(t)dt ≤C1w(x) for all balls B and almost all x ∈ B. (7.1.13)

It remains to understand what condition (7.1.13) really means. For every ball B, there

exists a null set N(B) such that (7.1.13) holds for all x in B\N(B). Let N be the union

of all the null sets N(B) for all balls B with centers in Qn and rational radii. Then

N is a null set and for every x in B \N, (7.1.13) holds for all balls B with centers in

Qn and rational radii. By density, (7.1.13) must also hold for all balls B that contain

a fixed x in Rn \N. It follows that for x ∈ Rn \N we have

M(w)(x) = supB∋x

1

|B|

∫

Bw(t)dt ≤C1w(x) . (7.1.14)

Therefore, assuming (7.1.8), we have arrived at the condition

M(w)(x)≤C1w(x) for almost all x ∈ Rn, (7.1.15)

where C1 is the same constant as in (7.1.13).

We later see that this deduction can be reversed and we can obtain (7.1.8) as a

consequence of (7.1.15). This motivates a careful study of condition (7.1.15), which

we refer to as the A1 condition. Since in all the previous arguments we could have

replaced balls with cubes, we give the following definitions in terms of cubes.

Definition 7.1.1. A function w(x)≥ 0 is called an A1 weight if

M(w)(x)≤C1w(x) for almost all x ∈ Rn (7.1.16)

for some constant C1. If w is an A1 weight, then the (finite) quantity

[w]A1= sup

Q cubes in Rn

(1

|Q|

∫

Qw(t)dt

)‖w−1‖L∞(Q) (7.1.17)

is called the A1 Muckenhoupt characteristic constant of w, or simply the A1 charac-

teristic constant of w. Note that A1 weights w satisfy

1

|Q|

∫

Qw(t)dt ≤ [w]A1

ess.infy∈Q

w(y) (7.1.18)

for all cubes Q in Rn.

Remark 7.1.2. We also define

[w]ballsA1

= supB balls in Rn

(1

|B|

∫

Bw(t)dt

)∥∥w−1∥∥

L∞(B). (7.1.19)

Using (7.1.13), we see that the smallest constant C1 that appears in (7.1.16) is equal

to the A1 characteristic constant of w as defined in (7.1.19). This is also equal to the

smallest constant that appears in (7.1.13). All these constants are bounded above and

below by dimensional multiples of [w]A1.

We now recall condition (7.1.5), which motivates the following definition of Ap

weights for 1 < p < ∞.

Definition 7.1.3. Let 1 < p < ∞. A weight w is said to be of class Ap if

supQ cubes in Rn

(1

|Q|

∫

Qw(x)dx

)(1

|Q|

∫

Qw(x)−

1p−1 dx

)p−1

< ∞ . (7.1.20)

The expression in (7.1.20) is called the Ap Muckenhoupt characteristic constant of

w (or simply the Ap characteristic constant of w) and is denoted by [w]Ap .

Remark 7.1.4. Note that Definitions 7.1.1 and 7.1.3 could have been given with the

set of all cubes in Rn replaced by the set of all balls in Rn. Defining [w]ballsAp

as in

(7.1.20) except that cubes are replaced by balls, we see that

(vn2−n

)p ≤[w]Ap

[w]ballsAp

≤(nn/2vn2−n

)p. (7.1.21)

7.1.2 Properties of Ap Weights

It is straightforward that translations, isotropic dilations, and scalar multiples of Ap

weights are also Ap weights with the same Ap characteristic. We summarize some

basic properties of Ap weights in the following proposition.

Proposition 7.1.5. Let w ∈ Ap for some 1≤ p < ∞. Then

(1) [δ λ (w)]Ap = [w]Ap , where δ λ (w)(x) = w(λx1, . . . ,λxn).

(2) [τz(w)]Ap = [w]Ap , where τz(w)(x) = w(x− z), z ∈ Rn.

(3) [λw]Ap = [w]Ap for all λ > 0.

(4) When 1 < p < ∞, the function w− 1

p−1 is in Ap′ with characteristic constant

[w− 1

p−1]

Ap′= [w]

1p−1

Ap.

Therefore, w ∈ A2 if and only if w−1 ∈ A2 and both weights have the same A2

characteristic constant.

(5) [w]Ap ≥ 1 for all w ∈ Ap. Equality holds if and only if w is a constant.

(6) The classes Ap are increasing as p increases; precisely, for 1 ≤ p < q < ∞ we

have

[w]Aq ≤ [w]Ap .

(7) limq→1+

[w]Aq = [w]A1if w ∈ A1.

(8) The following is an equivalent characterization of the Ap characteristic constant

of w:

[w]Ap = supQcubesin Rn

supf ∈ Lp(Q,wdt)∫

Q | f |pwdt>0

(1|Q|

∫Q | f (t)|dt

)p

1w(Q)

∫Q | f (t)|pw(t)dt

.

(9) The measure w(x)dx is doubling: precisely, for all λ > 1 and all cubes Q we

have

w(λQ)≤ λ np[w]Ap w(Q) .

(λQ denotes the cube with the same center as Q and side length λ times the side

length of Q.)

Proof. The simple proofs of (1), (2), and (3) are left as an exercise. Property (4) is

also easy to check and plays the role of duality in this context. To prove (5) we use

Holder’s inequality with exponents p and p′ to obtain

1 =1

|Q|

∫

Qdx =

1

|Q|

∫

Qw(x)

1p w(x)−

1p dx≤ [w]

1p

Ap,

with equality holding only when w(x)1p = cw(x)−

1p for some c > 0 (i.e., when w is a

constant). To prove (6), observe that 0 < q′−1 < p′−1≤ ∞ and that the statement

[w]Aq ≤ [w]Ap

is equivalent to the fact

∥∥w−1∥∥

Lq′−1(Q, dx|Q| )≤

∥∥w−1∥∥

Lp′−1(Q, dx|Q| )

.

Property (7) is a consequence of part (a) of Exercise 1.1.3.

To prove (8), apply Holder’s inequality with exponents p and p′ to get

(AvgQ

| f |)p =

(1

|Q|

∫

Q| f (x)|dx

)p

=

(1

|Q|

∫

Q| f (x)|w(x)

1p w(x)−

1p dx

)p

≤ 1

|Q|p(∫

Q| f (x)|pw(x)dx

)(∫

Qw(x)−

p′p dx

) p

p′

=

(1

ω(Q)

∫

Q| f (x)|pw(x)dx

)(1

|Q|

∫

Qw(x)dx

)(1

|Q|

∫

Qw(x)−

1p−1 dx

)p−1

≤ [w]Ap

(1

ω(Q)

∫

Q| f (x)|pw(x)dx

).

This argument proves the inequality ≥ in (8) when p > 1. In the case p = 1 the

obvious modification yields the same inequality. The reverse inequality follows by

taking f = (w+ ε)−p′/p as in (7.1.6) and letting ε → 0.

Applying (8) to the function f = χQ and putting λQ in the place of Q in (8), we

obtain

w(λQ)≤ λ np[w]Apw(Q) ,

which says that w(x)dx is a doubling measure. This proves (9).

Example 7.1.6. A positive measure dµ is called doubling if for some C < ∞,

µ(2B)≤Cµ(B) (7.1.22)

for all balls B. We show that the measures |x|a dx are doubling when a > −n. We

divide all balls B(x0,R) in Rn into two categories: balls of type I that satisfy |x0| ≥3R and type II that satisfy |x0|< 3R. For balls of type I we observe that

∫

B(x0,2R)|x|a dx ≤ vn(2R)n

(|x0|+2R)a when a≥ 0,

(|x0|−2R)a when a < 0,

∫

B(x0,R)|x|a dx ≥ vnRn

(|x0|−R)a when a≥ 0,

(|x0|+R)a when a < 0.

Since |x0| ≥ 3R, we have |x0|+2R≤ 4(|x0|−R) and |x0|−2R ≥ 14(|x0|+R), from

which (7.1.22) follows with C = 23n4|a|.For balls of type II, we have |x0| ≤ 3R and we note two things: first

∫

B(x0,2R)|x|a dx≤

∫

|x|≤5R|x|a dx = cnRn+a,

and second, since |x|a is radially decreasing for a < 0 and radially increasing for

a≥ 0, we have

∫

B(x0,R)|x|a dx≥

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∫

B(0,R)|x|a dx when a≥ 0,

∫

B(3Rx0|x0|

,R)|x|a dx when a < 0.

For x ∈ B(3Rx0|x0| ,R) we must have |x| ≥ 2R, and hence both integrals on the right

are at least a multiple of Rn+a. This establishes (7.1.22) for balls of type II.

Example 7.1.7. We investigate for which real numbers a the power function |x|a is

an Ap weight on Rn. For 1< p<∞, we examine for which a the following expression

is finite:

supB balls

(1

|B|

∫

B|x|a dx

)(1

|B|

∫

B|x|−a

p′p dx

) p

p′. (7.1.23)

As in the previous example we split the balls in Rn into those of type I and those of

type II. If B = B(x0,R) is of type I, then for x satisfying |x− x0| ≤ R we must have

2

3|x0| ≤ |x0|−R≤ |x| ≤ |x0|+R≤ 4

3|x0| ,

thus the expression inside the supremum in (7.1.23) is comparable to

|x0|a(|x0|−a

p′p) p

p′ = 1.

If B(x0,R) is a ball of type II, then B(0,5R) has size comparable to B(x0,R) and

contains it. Since the measure |x|a dx is doubling, the integrals of the function |x|aover B(x0,R) and over B(0,5R) are comparable. It suffices therefore to estimate the

expression inside the supremum in (7.1.23), in which we have replaced B(x0,R) by

B(0,5R). But this is

(1

vn(5R)n

∫

B(0,5R)|x|a dx

)(1

vn(5R)n

∫

B(0,5R)|x|−a

p′p dx

) p

p′

=

(n

(5R)n

∫ 5R

0ra+n−1dr

)(n

(5R)n

∫ 5R

0r−a

p′p +n−1

dr

) p

p′,

which is seen easily to be finite and independent of R exactly when −n < a < npp′ .

We conclude that |x|a is an Ap weight, 1 < p <∞, if and only if −n < a < n(p−1).The previous proof can be suitably modified to include the case p= 1. In this case

we obtain that |x|a is an A1 weight if and only if −n < a ≤ 0. As we have seen, the

measure |x|a dx is doubling on the larger range −n < a < ∞. Thus for a > n(p−1),the function |x|a provides an example of a doubling measure that is not in Ap.

Example 7.1.8. On Rn the function

u(x) =

log 1

|x| when |x|< 1e,

1 otherwise,

is an A1 weight. Indeed, to check condition (7.1.19) it suffices to consider balls of

type I and type II as defined in Example 7.1.6. In either case the required estimate

follows easily.

We now return to a point alluded to earlier, that the Ap condition implies the

boundedness of the Hardy–Littlewood maximal function M on the space Lp(w). To

this end we introduce four maximal functions acting on functions f that are locally

integrable with respect to w:

Mw( f )(x) = supB∋x

1

w(B)

∫

B| f |wdy ,

where the supremum is taken over open balls B that contain the point x and

Mw( f )(x) = sup

δ>0

1

w(B(x,δ ))

∫

B(x,δ )| f |wdy ,

Mwc ( f )(x) = sup

Q∋x

1

w(Q)

∫

Q| f |wdy ,

where Q is an open cube containing the point x, and

Mwc ( f )(x) = sup

δ>0

1

w(Q(x,δ ))

∫

Q(x,δ )| f |wdy ,

where Q(x,δ ) = ∏nj=1(x j − δ ,x j + δ ) is a cube of side length 2δ centered at

x = (x1, . . . ,xn). When w = 1, these maximal functions reduce to the standard ones

M( f ), M( f ), Mc( f ), and Mc( f ), the uncentered and centered Hardy–Littlewood

maximal functions with respect to balls and cubes, respectively.

Theorem 7.1.9. (a) Let w ∈ A1. Then we have

∥∥Mc

∥∥L1(w)→L1,∞(w)

≤ 3n[w]A1. (7.1.24)

(b) Let w ∈ Ap(Rn) for some 1 < p < ∞. Then there is a constant Cn,p such that

∥∥Mc

∥∥Lp(w)→Lp(w)

≤Cn,p[w]1

p−1

Ap. (7.1.25)

Since the operators Mc, Mc, M, and M are pointwise comparable, a similar conclu-

sions hold for the other three as well.

Proof. (a) Since dµ = wdx is a doubling measure and dµ(3Q) ≤ 3n[w]A1µ(Q),

using Proposition 7.1.5 (9) and Exercise 2.1.1 we obtain that Mwc maps L1(w) to

L1,∞(w) with norm at most 3n[w]A1. This proves (7.1.24).

(b) Fix a weight w in Ap and let σ = w− 1

p−1 be its dual weight. Fix an open cube

Q = Q(x0,r) in Rn with center x0 and side length 2r and write

1

|Q|

∫

Q| f |dy =

w(Q)1

p−1σ(3Q)

|Q|p

p−1

|Q|w(Q)

(1

σ(3Q)

∫

Q| f |dy

)p−1 1p−1

. (7.1.26)

For any x ∈Q, consider the cube Q(x,2r). Then Q Q(x,2r) 3Q = Q(x0,3r) and

thus1

σ(3Q)

∫

Q| f |dy≤ 1

σ(Q(x,2r))

∫

Q(x,2r)| f |dy≤M

σc (| f |σ−1)(x)

for any x ∈ Q. Inserting this expression in (7.1.26), we obtain

1

|Q|

∫

Q| f |dy≤ w(Q)

1p−1σ(3Q)

|Q|p

p−1

1

w(Q)

∫

QM

σc (| f |σ−1)p−1 dy

1p−1

. (7.1.27)

Since one may easily verify that

w(Q)σ(3Q)p−1

|Q|p ≤ 3np[w]Ap ,

it follows that

1

|Q|

∫

Q| f |dy≤ 3

npp−1 [w]

1p−1

Ap

(M

wc

[(M

σc (| f |σ−1)

)p−1w−1

](x0)

) 1p−1

,

since x0 is the center of Q. Hence, we have

Mc( f )≤ 3np

p−1 [w]1

p−1

Ap

(M

wc

[(M

σc (| f |σ−1)

)p−1w−1

]) 1p−1

.

Applying Lp(w) norms, we deduce

∥∥Mc( f )∥∥

Lp(w)≤ 3

npp−1 [w]

1p−1

Ap

∥∥Mwc

[(M

σc (| f |σ−1)

)p−1w−1

]∥∥ 1p−1

Lp′ (w)

≤ 3np

p−1 [w]1

p−1

Ap

∥∥Mwc

∥∥ 1p−1

Lp′ (w)→Lp′ (w)

∥∥(Mσc (| f |σ−1)

)p−1w−1

∥∥ 1p−1

Lp′ (w)

= 3np

p−1 [w]1

p−1

Ap

∥∥Mwc

∥∥ 1p−1


∥∥Mσc (| f |σ−1)

∥∥Lp(σ)

≤ 3np

p−1 [w]1

p−1

Ap

∥∥Mwc

∥∥ 1p−1


∥∥Mσc

∥∥Lp(σ)→Lp(σ)

∥∥ f∥∥

Lp(w),

and conclusion (7.1.25) follows, provided we show that

∥∥Mwc

∥∥Lq(w)→Lq(w)

≤C(q,n)< ∞ (7.1.28)

for any 1 < q < ∞ and any weight w.

We obtain this estimate by interpolation. Obviously (7.1.28) is valid when q = ∞with C(∞,n) = 1. If we prove that

∥∥Mwc

∥∥L1(w)→L1,∞(w)

≤C(1,n)< ∞ , (7.1.29)

then (7.1.28) will follow from Theorem 1.3.2.

To prove (7.1.29) we fix f ∈ L1(Rn,wdx). We first show that the set

Eλ = Mwc ( f )> λ

is open. For any r > 0, let Q(x,r) denote an open cube of side length 2r with center

x ∈ Rn. If we show that for any r > 0 and x ∈ Rn the function

x → 1

w(Q(x,r))

∫

Q(x,r)| f |wdy (7.1.30)

is continuous, then Mwc ( f ) is the supremum of continuous functions; hence it is

lower semicontinuous and thus the set Eλ is open. But this is straightforward. If

xn → x0, then w(Q(xn,r))→ w(Q(x0,r)) and also∫

Q(xn,r)| f |wdy→ ∫

Q(x0,r)| f |wdy

by the Lebesgue dominated convergence theorem. Since w(Q(x0,r)) = 0, it follows

that the function in (7.1.30) is continuous.

Given K a compact subset of Eλ , for any x ∈ K select an open cube Qx centered

at x such that1

w(Qx)

∫

Qx

| f |wdy > λ .

Applying Lemma 7.1.10 (proved immediately afterward) we find a subfamily

Qx jm

j=1 of the family of the balls Qx : x ∈ K such that (7.1.31) and (7.1.32)

hold. Then

w(K)≤m

∑j=1

w(Qx j)≤

m

∑j=1

1

λ

∫

Qx j

| f |wdy≤ 24n

λ

∫

Rn| f |wdy ,

where the last inequality follows by multiplying (7.1.32) by | f |w and integrating

over Rn. Taking the supremum over all compact subsets K of Eλ and using the inner

regularity of wdx, which is a consequence of the Lebesgue monotone convergence

theorem, we deduce that Mwc maps L1(w) to L1,∞(w) with constant at most 24n. Thus

(7.1.29) holds with C(1,n) = 24n.

Lemma 7.1.10. Let K be a bounded set in Rn and for every x ∈ K, let Qx be an open

cube with center x and sides parallel to the axes. Then there are an m ∈ Z+ ∪∞and a sequence of points x jm

j=1 in K such that

K m⋃

j=1

Qx j(7.1.31)

and for almost all y ∈ Rn one has

m

∑j=1

χQx j(y)≤ 24n . (7.1.32)

Proof. Let s0 = supℓ(Qx) : x ∈ K. If s0 = ∞, then there exists x1 ∈ K such that

ℓ(Qx1) > 4L, where [−L,L]n contains K. Then K is contained in Qx1

and the state-

ment of the lemma is valid with m = 1.

Suppose now that s0 < ∞. Select x1 ∈ K such that ℓ(Qx1)> s0/2. Then define

K1 = K \Qx1, s1 = supℓ(Qx) : x ∈ K1 ,

and select x2 ∈ K1 such that ℓ(Qx2)> s1/2. Next define

K2 = K \ (Qx1∪Qx2

) , s2 = supℓ(Qx) : x ∈ K2 ,

and select x3 ∈ K2 such that ℓ(Qx3) > s2/2. Continue until the first integer m is

found such that Km is an empty set. If no such integer exists, continue this process

indefinitely and set m = ∞.

We claim that for all i = j we have 13Qxi∩ 1

3Qx j

= /0. Indeed, suppose that i > j.

Then xi ∈ Ki−1 = K \ (Qx1∪·· ·∪Qxi−1

); thus xi /∈Q j. Also xi ∈ Ki−1 K j−1, which

implies that ℓ(Qxi) ≤ s j−1 < 2ℓ(Qx j

). If xi /∈ Q j and ℓ(Qx j) > 1

2ℓ(Qxi

), it easily

follows that 13Qxi∩ 1

3Qx j

= /0.

We now prove (7.1.31). If m < ∞, then Km = /0 and therefore K ⋃m

j=1 Qx j. If

m = ∞, then there is an infinite number of selected cubes Qx j. Since the cubes 1

3Qx j

are pairwise disjoint and have centers in a bounded set, it must be the case that

some subsequence of the sequence of their lengths converges to zero. If there exists

a y ∈ K \⋃∞j=1 Qx j

, this y would belong to all K j, j = 1,2, . . . , and then s j ≥ ℓ(Qy)for all j. Since some subsequence of the s j’s tends to zero, it would follow that

ℓ(Qy) = 0, which would force the open cube Qy to be the empty set, a contradiction.

Thus (7.1.31) holds.

Finally, we show that ∑mj=1 χQx j

(y)≤ 24n for almost every point y ∈Rn. To prove

this we consider the n hyperplanes Hi that are parallel to the coordinate hyperplanes

and pass through the point y. Then we write Rn as a union of n hyperplanes Hi of

n-dimensional Lebesgue measure zero and 2n higher-dimensional open “octants”

Or, henceforth called orthants. We fix a y ∈ Rn and we show that there are only

12n points x j such that y lies in Or ∩Qx jfor a given open orthant Or. To prove this

assertion, setting |z|ℓ∞ = sup1≤i≤n |zi| for points z = (z1, . . . ,zn) in Rn, we pick an

xk0∈ K ∩Or such that Qxk0

contains y and |xk0− y|ℓ∞ is the largest possible among

all |x j−y|ℓ∞ . If x j is another point in K∩Or such that Qx jcontains y, then we claim

that x j ∈ Qxk0. Indeed, to show this we notice that for each i ∈ 1, . . . ,n we have

|x j,i− xk0,i| =∣∣x j,i− yi− (xk0,i− yi)

∣∣=

∣∣|x j,i− yi|− |xk0,i− yi|∣∣

≤ max(|xk0,i− yi|, |x j,i− yi|

)

≤ max(|xk0

− y|ℓ∞ , |x j− y|ℓ∞)

= |xk0− y|ℓ∞

< 12ℓ(Qxk0

) ,

where the second equality is due to the fact that x j,xk0lie in the same orthant and

the last inequality in the fact that y ∈ Qxk0; it follows that x j lies in Qxk0

.

We observed previously that i > j implies xi /∈Q j. Since x j lies in Qxk0, one must

then have j ≤ k0, which implies that 12ℓ(Qxk0

) < ℓ(Qx j). Thus all cubes Qx j

with

centers in K∩Or that contain the fixed point y have side lengths comparable to that

of Qxk0. A simple geometric argument now gives that there are at most finitely many

cubes Qx jof side length between α and 2α that contain the given point y such that

13Qx j

are pairwise disjoint. Indeed, let α = 12ℓ(Qxk0

) and let Qxrr∈I be the cubes

with these properties. Then we have

αn|I|3n

≤∑r∈I

∣∣ 13Qxr

∣∣=∣∣⋃

r∈I

13Qxr

∣∣≤∣∣⋃

r∈I

Qxr

∣∣≤ (4α)n ,

since all the cubes Qxr contain the point y and have length at most 2α and they

must therefore be contained in a cube of side length 4α centered at y. This obser-

vation shows that |I| ≤ 12n, and since there are 2n sets Or, we conclude the proof

of (7.1.32).

Remark 7.1.11. Without use of the covering Lemma 7.1.10, (7.1.29) can be proved

via the doubling property of w (cf. Exercise 2.1.1(a)), but then the resulting constant

C(q,n) would depend on the doubling constant of the measure wdx and thus on

[w]Ap ; this would yield a worse dependence on [w]Ap in the constant in (7.1.25).

Exercises

7.1.1. Let k be a nonnegative measurable function such that k,k−1 are in L∞(Rn).Prove that if w is an Ap weight for some 1≤ p < ∞, then so is kw.

7.1.2. Let w1, w2 be two A1 weights and let 1 < p < ∞. Prove that w1w1−p2 is an Ap

weight by showing that

[w1w1−p2 ]Ap ≤ [w1]A1

[w2]p−1A1

.

7.1.3. Suppose that w ∈ Ap for some p ∈ [1,∞) and 0 < δ < 1. Prove that wδ ∈ Aq,

where q = δ p+1−δ , by showing that

[wδ ]Aq ≤ [w]δAp.

7.1.4. Show that if the Ap characteristic constants of a weight w are uniformly

bounded for all p > 1, then w ∈ A1.

7.1.5. Let w0 ∈ Ap0and w1 ∈ Ap1

for some 1≤ p0, p1 <∞. Let 0≤ θ ≤ 1 and define

1

p=

1−θp0

+θ

p1and w

1p = w

1−θp0

0 wθp11 .

Prove that

[w]Ap ≤ [w0](1−θ) p

p0Ap0

[w1]θ p

p1Ap1

;

thus w is in Ap.

7.1.6. ([122]) Fix 1 < p < ∞. A pair of weights (u,w) that satisfies

[u,w](Ap,Ap) = supQ cubes

in Rn

(1

|Q|

∫

Qudx

)(1

|Q|

∫

Qw− 1

p−1 dx

)p−1

< ∞

is said to be of class (Ap,Ap). The quantity [u,w](Ap,Ap) is called the (Ap,Ap) char-

acteristic constant of the pair (u,w).(a) Suppose that pair of weights (u,w) is of class (Ap,Ap). Show that for all non-

negative measurable functions f and all cubes Q′ we have

(1

|Q′|

∫

Q′| f |dx

)p

u(Q′)≤C0

∫

Q′| f |pwdx ,

where C0 = [u,w](Ap,Ap).

(b) Suppose that a pair of weights (u,w) satisfies the inequality in part (a) for some

constant C0. Prove that M maps Lp(w) to Lp,∞(u) with norm at most C(n, p)C1/p

0 ,

where C(n, p) is a fixed constant.

(c) Suppose that for a pair of weights (u,w), M maps Lp(w) to Lp,∞(u). Show that

the pair (u,w) is of class (Ap,Ap).[Hint: Part (b): Replacing f by f χQ in part (a), where Q⊆ Q′, obtain that

u(Q′)≤C0|Q′|p∫

Q | f |pwdx(∫

Q | f |dx)p .

Then use Exercise 5.3.9 to find disjoint cubes Q j such that the set Eα = x ∈ Rn :

Mc( f )(x) > α is contained in the union of 3Q j and α4n < 1

|Q j |∫

Q j| f (t)|dt ≤ α

2n .

Then u(Eα)≤∑ j u(3Q j), and bound each u(3Q j) by taking Q′= 3Q j and Q=Q j in

the preceding estimate. Part (c): First prove the assertion in part (b) and then derive

the inequality in part (a) by adapting the idea in the discussion in the beginning of

Subsection 7.1.1.]

7.1.7. ([122]) Let 1 < p < ∞ and let (u,w) be a pair of weights of class (Ap,Ap).Show that for any q with p < q < ∞ there is a constant Cp,q,n < ∞ such that for all

f ∈ Lq(w) we have

(∫

RnM( f )(x)qu(x)dx

)1/q

≤Cp,q,n

(∫

Rnf (x)qw(x)dx

)1/q

.

[Hint: Use Exercise 7.1.6 and interpolate between Lp and L∞.

]

7.1.8. Let k > 0. For an A1 weight w show that [min(w,k)]A1≤ [w]A1

. If 1 < p < ∞and w ∈ Ap, show that

[min(w,k)]Ap ≤ cp[w]Ap ,

where cp = 1 if 1 < p≤ 2 and cp = 2p−1 if 2 < p < ∞.[Hint: Use the inequality 1

|Q|∫

Q min(w,k)−1

p−1 dx≤ 1|Q|

∫Q w

− 1p−1 dx+k

− 1p−1 and also

1|Q|

∫Q min(w,k)dx≤min

k, 1|Q|

∫Q wdx

.]

7.1.9. Suppose that w j ∈ Ap jwith 1 ≤ j ≤ m for some 1 ≤ p1, . . . , pm < ∞ and let

0 < θ1, . . . ,θm < 1 be such that θ1 + · · ·+θm = 1. Show that

wθ11 · · ·wθm

m ∈ Amaxp1,...,pm .

[Hint: First note that each weight w j lies in Amaxp1,...,pm and then apply Holder’s

inequality.]

7.1.10. Let w1 ∈ Ap1and w2 ∈ Ap2

for some 1≤ p1, p2 < ∞. Prove that

[w1 +w2]Ap ≤ [w1]Ap1+[w2]Ap2

,

where p = max(p1, p2).

7.1.11. Show that the function

u(x) =

log 1

|x| when |x|< 1e,

1 otherwise,

in Example 7.1.8 is an A1 weight on Rn.[Hint: Use [u]balls

A1instead of [u]A1

and consider balls of type I and II as in Example

7.1.7.]

7.1.12. Let 1 λ)≤3n([w]balls

A1)2

λ

∫

M(g)>λ|g|wdx

and then use the characterization of Lp,∞ given in Exercise 1.1.12.]

7.2 Reverse Holder Inequality for Ap Weights and Consequences

An essential property of Ap weights is that they assign to subsets of balls mass

proportional to the percentage of the Lebesgue measure of the subset within the

ball. The following lemma provides a way to quantify this statement.

Lemma 7.2.1. Let w ∈ Ap for some 1≤ p < ∞ and let 0 < α < 1. Then there exists

β < 1 such that whenever S is a measurable subset of a cube Q that satisfies |S| ≤α|Q|, we have w(S)≤ β w(Q).

Proof. Taking f = χA in property (8) of Proposition 7.1.5, we obtain

( |A||Q|

)p

≤ [w]Ap

w(A)

w(Q). (7.2.1)

We write S = Q\A to get

(1− |S||Q|

)p

≤ [w]Ap

(1− w(S)

w(Q)

). (7.2.2)

Given 0 < α < 1, set

β = 1− (1−α)p

[w]Ap

(7.2.3)

and use (7.2.2) to obtain the required conclusion.

7.2.1 The Reverse Holder Property of Ap Weights

We are now ready to state and prove one of the main results of the theory of weights,

the reverse Holder inequality for Ap weights.

Theorem 7.2.2. Let w ∈ Ap for some 1 ≤ p < ∞. Then there exist constants C and

γ > 0 that depend only on the dimension n, on p, and on [w]Ap such that for every

cube Q we have

(1

|Q|

∫

Qw(t)1+γ dt

) 11+γ

≤ C

|Q|

∫

Qw(t)dt . (7.2.4)

Proof. Let us fix a cube Q and set

α0 =1

|Q|

∫

Qw(x)dx .

We also fix 0 < α < 1. We define an increasing sequence of scalars

α0 < α1 < α2 < · · ·< αk < · · ·

7.2 Reverse Holder Inequality and Consequences 515

for k ≥ 0 by setting

αk+1 = 2nα−1αk or αk = (2nα−1)kα0,

and for each k ≥ 1 we apply a Calderon–Zygmund decomposition to w at height αk.

Precisely, for dyadic subcubes R of Q, we let

1

|R|

∫

Rw(x)dx > αk (7.2.5)

be the selection criterion. Since Q does not satisfy the selection criterion, it is not

selected. We divide the cube Q into a mesh of 2n subcubes of equal side length, and

among these cubes we select those that satisfy (7.2.5). We subdivide each unselected

subcube into 2n cubes of equal side length and we continue in this way indefinitely.

We denote by Qk, j j the collection of all selected subcubes of Q. We observe that

the following properties are satisfied:

(1) αk <1

|Qk, j|

∫

Qk, j

w(t)dt ≤ 2nαk.

(2) For almost all x /∈Uk we have w(x)≤ αk, where Uk =⋃j

Qk, j.

(3) Each Qk+1, j is contained in some Qk,l .

Property (1) is satisfied since the unique dyadic parent of Qk, j was not chosen in the

selection procedure. Property (2) follows from the Lebesgue differentiation theorem

using the fact that for almost all x /∈Uk there exists a sequence of unselected cubes

of decreasing lengths whose closures’ intersection is the singleton x. Property (3)

is satisfied since each Qk, j is the maximal subcube of Q satisfying (7.2.5). And since

the average of w over Qk+1, j is also bigger than αk, it follows that Qk+1, j must be

contained in some maximal cube that possesses this property.

We now compute the portion of Qk,l that is covered by cubes of the form Qk+1, j

for some j. We have

2nαk ≥1

|Qk,l |

∫

Qk,l∩Uk+1

w(t)dt

=1

|Qk,l | ∑j:Qk+1, j⊆Qk,l

|Qk+1, j|1

|Qk+1, j|

∫

Qk+1, j

w(t)dt

>

∣∣Qk,l ∩Uk+1

∣∣|Qk,l |

αk+1

=

∣∣Qk,l ∩Uk+1

∣∣|Qk,l |

2nα−1αk .

It follows that∣∣Qk,l ∩Uk+1

∣∣≤ α|Qk,l |; thus, applying Lemma 7.2.1, we obtain

w(Qk,l ∩Uk+1)

w(Qk,l)< β = 1− (1−α)p

[w]Ap

,

from which, summing over all l, we obtain

w(Uk+1)≤ βw(Uk) .

The latter gives w(Uk)≤ β kw(U0). We also have |Uk+1| ≤ α|Uk|; hence |Uk| → 0 as

k→∞. Therefore, the intersection of the Uk’s is a set of Lebesgue measure zero. We

can therefore write

Q =(Q\U0

)⋃( ∞⋃

k=0

Uk \Uk+1

)

modulo a set of Lebesgue measure zero. Let us now find a γ > 0 such that the reverse

Holder inequality (7.2.4) holds. We have w(x) ≤ αk for almost all x in Q \Uk and

therefore

∫

Qw(t)1+γ dt =

∫

Q\U0

w(t)γw(t)dt +∞

∑k=0

∫

Uk\Uk+1

w(t)γw(t)dt

≤ αγ0 w(Q\U0)+

∞

∑k=0

αγk+1w(Uk)

≤ αγ0 w(Q\U0)+

∞

∑k=0

((2nα−1)k+1α0)γβ kw(U0)

≤ αγ0

(1+(2nα−1)γ

∞

∑k=0

(2nα−1)γkβ k

)w(Q)

=

(1

|Q|

∫

Qw(t)dt

)γ(1+

(2nα−1)γ

1− (2nα−1)γβ

)∫

Qw(t)dt ,

provided γ > 0 is chosen small enough that (2nα−1)γβ < 1. Keeping track of the

constants, we conclude the proof of the theorem with

γ =1

2

− logβ

log2n− logα=

log([w]Ap

)− log

([w]Ap − (1−α)p

)

2log 2n

α

(7.2.6)

and

Cγ+1 = 1+(2nα−1)γ

1− (2nα−1)γβ

= 1+(2nα−1)γ

1− (2nα−1)γ(1− (1−α)p

[w]Ap

)

= 1+1

(2nα−1)−γ −(1− (1−α)p

[w]Ap

) ,

which yields

C =

[1+

1(1− (1−α)p

[w]Ap

) 12 −

(1− (1−α)p

[w]Ap

)] 2log 2n

α

2log 2nα −log

(1− (1−α)p

[w]Ap

). (7.2.7)

Note that up to this point, α was an arbitrary number in (0,1).

Remark 7.2.3. It is worth observing that for α such that (1−α)p = 34, the constant

γ in (7.2.6) decreases as [w]Ap increases, while the constant C in (7.2.7) increases as

[w]Ap increases. This is because 1− 34[w]−1

Ap≥ 1

4and for t ∈ ( 1

4,1) the function

√t−t

is decreasing. This allows us to obtain the following stronger version of Theorem

7.2.2: For any 1 ≤ p < ∞ and B > 1, there exist positive constants C = C(n, p,B)and γ = γ(n, p,B) such that for all w ∈ Ap satisfying [w]Ap ≤ B the reverse Holder

condition (7.2.4) holds for every cube Q. See Exercise 7.2.4(a) for details.

Observe that in the proof of Theorem 7.2.2 it was crucial to know that for some

0 < α,β < 1 we have

|S| ≤ α |Q| =⇒ w(S)≤ β w(Q) (7.2.8)

whenever S is a subset of the cube Q. No special property of Lebesgue measure was

used in the proof of Theorem 7.2.2 other than its doubling property. Therefore, it is

reasonable to ask whether Lebesgue measure in (7.2.8) can be replaced by a general

measure µ satisfying the doubling property

µ(3Q)≤Cn µ(Q)< ∞ (7.2.9)

for all cubes Q in Rn. A straightforward adjustment of the proof of the previous

theorem indicates that this is indeed the case.

Corollary 7.2.4. Let w be a weight and let µ be a measure on Rn satisfying (7.2.9).

Suppose that there exist 0 < α,β < 1, such that

µ(S)≤ α µ(Q) =⇒∫

Sw(t)dµ(t)≤ β

∫

Qw(t)dµ(t)

whenever S is a µ-measurable subset of a cube Q. Then there exist 0 < C,γ < ∞[which depend only on the dimension n, the constant Cn in (7.2.9), α , and β ] such

that for every cube Q in Rn we have

(1

µ(Q)

∫

Qw(t)1+γ dµ(t)

) 11+γ

≤ C

µ(Q)

∫

Qw(t)dµ(t). (7.2.10)

Proof. The proof of the corollary can be obtained almost verbatim from that of

Theorem 7.2.2 by replacing Lebesgue measure with the doubling measure dµ and

the constant 2n by Cn.

Precisely, we define αk = (Cnα−1)kα0, where α0 is the µ-average of w over Q;

then properties (1), (2), (3) concerning the selected cubes Qk, j j are replaced by

(1µ ) αk <1

µ(Qk, j)

∫

Qk, j

w(t)dµ(t)≤Cnαk.

(2µ ) On Q\Uk we have w≤ αk µ-almost everywhere, where Uk =⋃j

Qk, j.

(3µ ) Each Qk+1, j is contained in some Qk,l .

To prove the upper inequality in (1µ ) we use that the dyadic parent of each selected

cube Qk, j was not selected and is contained in 3Qk, j. To prove (2µ ) we need a dif-

ferentiation theorem for doubling measures, analogous to that in Corollary 2.1.16.

This can be found in Exercise 2.1.1. The remaining details of the proof are trivially

adapted to the new setting. The conclusion is that for

0 < γ <− logβ

logCn− logα(7.2.11)

and

C =

[1+

(Cnα−1)γ

1− (Cnα−1)γβ

] 1γ+1

, (7.2.12)

(7.2.10) is satisfied. Notice that the choice of the constants (7.2.6) and (7.2.7) is valid

in this case with Cn in place of 2n.

7.2.2 Consequences of the Reverse Holder Property

Having established the crucial reverse Holder inequality for Ap weights, we now

pass to some very important applications. Among them, the first result of this section

yields that an Ap weight that lies a priori in L1loc(R

n) must actually lie in the better

space L1+σloc (Rn) for some σ > 0 depending on the weight.

Theorem 7.2.5. If w ∈ Ap for some 1 ≤ p < ∞, then there exists a number γ > 0

(that depends on n, p, and [w]Ap) such that w1+γ ∈ Ap.

Proof. Let C be the constant in the proof of Theorem 7.2.2. When p = 1, we apply

the reverse Holder inequality of Theorem 7.2.2 to the weight w to obtain

1

|Q|

∫

Qw(t)1+γ dt ≤

(C

|Q|

∫

Qw(t)dt

)1+γ

≤C1+γ [w]1+γA1

w(x)1+γ

for almost all x in the cube Q. Therefore, w1+γ is an A1 weight with characteristic

constant at most C1+γ [w]1+γA1

. When p > 1, there exist γ1,γ2 > 0 and C1,C2 > 0 such

that the reverse Holder inequality of Theorem 7.2.2 holds for the weights w ∈ Ap

and w− 1

p−1 ∈ Ap′ , that is,

(1

|Q|

∫

Qw(t)1+γ1dt

) 11+γ1 ≤ C1

|Q|

∫

Qw(t)dt,

(1

|Q|

∫

Qw(t)−

1p−1 (1+γ2)dt

) 11+γ2 ≤ C2

|Q|

∫

Qw(t)−

1p−1 dt .

Taking γ = min(γ1,γ2), both inequalities are satisfied with γ in the place of γ1, γ2. It

follows that w1+γ is in Ap and satisfies

[w1+γ ]Ap ≤ (C1Cp−12 )1+γ [w]

1+γAp

. (7.2.13)

This concludes the proof of the theorem.

Corollary 7.2.6. For any 1< p<∞ and for every w∈Ap there is a q= q(n, p, [w]Ap)with q < p such that w ∈ Aq. In other words, we have

Ap =⋃

q∈(1,p)Aq .

Proof. Given w ∈ Ap, let γ ,C1,C2 be as in the proof of Theorem 7.2.5. In view of

the result in Exercise 7.1.3 with δ = 1/(1+ γ), if w1+γ ∈ Ap and

q = p1

1+ γ+1− 1

1+ γ=

p+ γ

1+ γ,

then w ∈ Aq and

[w]Aq = [(w1+γ)1

1+γ ]Aq ≤[w1+γ

] 11+γ

Ap≤C1C

p−12 [w]Ap ,

where the last estimate comes from (7.2.13). Since 1 < q = p+γ1+γ < p, the required

conclusion follows. Observe that the constants C1Cp−12 , q, and 1

γ increase as [w]Ap

increases.

Another powerful consequence of the reverse Holder property of Ap weights is

the following characterization of all A1 weights.

Theorem 7.2.7. Let w be an A1 weight. Then there exist 0 < ε < 1, a nonnegative

function k such that k,k−1 ∈ L∞, and a nonnegative locally integrable function f

that satisfies M( f )< ∞ a.e. such that

w(x) = k(x)M( f )(x)ε . (7.2.14)

Conversely, given a nonnegative function k such that k,k−1 ∈ L∞ and given a

nonnegative locally integrable function f that satisfies M( f ) < ∞ a.e., define w via

(7.2.14). Then w is an A1 weight that satisfies

[w]A1≤ Cn

1− ε ‖k‖L∞‖k−1‖L∞ , (7.2.15)

where Cn is a universal dimensional constant.

Proof. In view of Theorem 7.2.2, there exist 0 < γ ,C < ∞ such that the reverse

Holder condition

(1

|Q|

∫

Qw(t)1+γ dt

) 11+γ

≤ C

|Q|

∫

Qw(t)dt ≤C [w]A1

w(x) (7.2.16)

holds for all cubes Q and for all x in Q\EQ, where EQ is a null subset of Q. We set

ε =1

1+ γand f (x) = w(x)1+γ = w(x)

1ε .

Letting N be the union of EQ over all Q with rational radii and centers in Qn, it

follows from (7.2.16) that the uncentered Hardy–Littlewood maximal function Mc( f )with respect to cubes satisfies

Mc( f )(x)≤C1+γ [w]1+γA1

f (x) for x ∈ Rn \N.

This implies that M( f ) ≤ CnC1+γ [w]1+γA1

f a.e. for some constant Cn that depends

only on the dimension. We now set

k(x) =f (x)ε

M( f )(x)ε,

and we observe that C−1C−εn [w]−1A1≤ k ≤ 1 a.e.

It remains to prove the converse. Given a weight w = kM( f )ε in the form (7.2.14)

and a cube Q, it suffices to show that

1

|Q|

∫

QM( f )(t)ε dt ≤ Cn

1− εM( f )ε(x) for almost all x ∈ Q, (7.2.17)

since then (7.2.15) follows trivially from (7.2.17) with w = kM( f )ε using that

k,k−1 ∈ L∞. To prove (7.2.17), we write

f = f χ3Q + f χ(3Q)c .

Then

1

|Q|

∫

QM( f χ3Q)(t)

ε dt ≤ C′n1− ε

(1

|Q|

∫

Rn( f χ3Q)(t)dt

)ε

(7.2.18)

in view of Kolmogorov’s inequality (Exercise 2.1.5). But the last expression in

(7.2.18) is at most a dimensional multiple of M( f )(x)ε for almost all x ∈ Q, which

proves (7.2.17) when f is replaced by f χ3Q on the left-hand side of the inequality.

And for f χ(3Q)c we only need to notice that

M( f χ(3Q)c)(t)≤ 2nM( f χ(3Q)c)(t)≤ 2nn

n2 M( f )(x)

for all x, t in Q, since any ball B centered at t that gives a nonzero average for

f χ(3Q)c must have radius at least the side length of Q, and thus√

nB must also

contain x. (Here M is the centered Hardy–Littlewood maximal operator introduced

in Definition 2.1.1.) Hence (7.2.17) also holds when f is replaced by f χ(3Q)c on the

left-hand side. Combining these two estimates and using the subadditivity property

M( f1 + f2)ε ≤M( f1)

ε +M( f2)ε , we obtain (7.2.17).

We end this section with the following consequence of the reverse Holder prop-

erty of Ap weights which can be viewed as a reverse property to (7.2.1).

Proposition 7.2.8. Let 1≤ p <∞ and w ∈ Ap. Then there exist δ ∈ (0,1) and C > 0

depending only on n, p, and [w]Ap such that for any cube Q and any measurable

subset S of Q we havew(S)

w(Q)≤C

( |S||Q|

)δ.

Proof. Let C and γ be as in Theorem 7.2.2. We use Holder’s inequality to write

w(S)

w(Q)=

1

w(Q)

∫

Qw(x)χS(x)dx

≤ 1

w(Q)

(∫

Qw(x)1+γ dx

) 11+γ

|S|γ

1+γ

=1

w(Q)

(1

|Q|

∫

Qw(x)1+γ dx

) 11+γ

|Q|1

1+γ |S|γ

1+γ

=C

w(Q)

(∫

Qw(x)dx

)|Q|−

γ1+γ |S|

γ1+γ

= C( |S||Q|

)δ,

where δ = γ1+γ . This proves the assertion.

Exercises

7.2.1. Let w ∈ Ap for some 1 < p < ∞ and let 1 ≤ q < ∞. Prove that the sublinear

operator

S( f ) =(M(| f |qw)w−1

) 1q

is bounded on Lp′q(w).

7.2.2. Let v be a real-valued locally integrable function on Rn and let 1 < p < ∞.

For a cube Q, let νQ be the average of ν over Q.

(a) If ev is an Ap weight, show that

supQ cubes

1

|Q|

∫

Qev(t)−vQ dt ≤ [eν ]Ap ,

supQ cubes

1

|Q|

∫

Qe−(v(t)−vQ)

1p−1 dt ≤ [eν ]Ap .

(b) Conversely, if the preceding inequalities hold with some constant C in place of

[ν ]Ap , then ν lies in Ap with [ν ]Ap ≤C.[Hint: Part (a): If ev ∈ Ap, use that

1

|Q|

∫

Qev(t)−vQ dt ≤

(Avg

Q

e− v

p−1

)p−1 (Avg

Q

ev)

and obtain a similar estimate for the second quantity.]

7.2.3. This exercise assumes familiarity with the space BMO.

(a) Show that if ϕ ∈ A2, then logϕ ∈ BMO and ‖ logϕ‖BMO ≤ [ϕ]A2.

(b) Prove that every BMO function is equal to a constant multiple of the logarithm

of an A2 weight. Precisely, given f ∈ BMO show that

[ec f

]A2≤ 1+2e ,

where c = 1/(2n+1‖ f‖BMO).(c) Prove that if ϕ is in Ap for some 1 < p < ∞, then logϕ is in BMO by showing

that

∥∥ logϕ∥∥

BMO≤

⎧⎨⎩[ϕ]Ap when 1 < p≤ 2,

(p−1)[ϕ]1

p−1

Apwhen 2 2.]

7.2.4. Prove the following quantitative versions of Theorem 7.2.2 and

Corollary 7.2.6.

(a) For any 1 ≤ p < ∞ and B > 1, there exists a positive constant C3(n, p,B) and

γ = γ(n, p,B) such that for all w ∈ Ap satisfying [w]Ap ≤ B, (7.2.4) holds for every

cube Q with C3(n, p,B) in place of C.

(b) Given any 1 1 there exists a constant C4(n, p,B) and δ =δ (n, p,B) such that for all w ∈ Ap we have

[w]Ap ≤ B =⇒ [w]Ap−δ ≤C4(n, p,B) .

7.2.5. Given a positive doubling measure µ on Rn, define the characteristic constant

[w]Ap(µ) and the class Ap(µ) for 1 < p < ∞.

(a) Show that statement (8) of Proposition 7.1.5 remains valid if Lebesgue measure

is replaced by µ .

(b) Obtain as a consequence that if w ∈ Ap(µ), then for all cubes Q and all µ-

measurable subsets A of Q we have

(µ(A)

µ(Q)

)p

≤ [w]Ap(µ)w(A)

w(Q).

Conclude that if Lebesgue measure is replaced by µ in Lemma 7.2.1, then the lemma

is valid for w ∈ Ap(µ).(c) Use Corollary 7.2.4 to obtain that weights in Ap(µ) satisfy a reverse Holder

condition.

(d) Prove that given a weight w ∈ Ap(µ), there exists 1 < q < p, which depends on

[w]Ap(µ), such that w ∈ Aq(µ).

7.2.6. Let 1< q<∞ and µ a positive measure on Rn. We say that a positive function

K on Rn satisfies a reverse Holder condition of order q with respect to µ , symboli-

cally K ∈ RHq(µ), if

[K]RHq(µ) = supQcubes in Rn

(1

µ(Q)

∫Q Kq dµ

) 1q

1µ(Q)

∫Q K dµ

< ∞ .

For positive functions u,v on Rn and 1 < p < ∞, show that

[vu−1]RHp′ (udx) = [uv−1]1p

Ap(vdx),

that is, vu−1 satisfies a reverse Holder condition of order p′ with respect to udx if

and only if uv−1 is in Ap(vdx). Conclude that

w ∈ RHp′(dx) ⇐⇒ w−1 ∈ Ap(wdx) ,

w ∈ Ap(dx) ⇐⇒ w−1 ∈ RHp′(wdx) .

7.2.7. ([125]) Suppose that a positive function K on Rn lies in RHp(dx) for some

1 0 such that K lies in RHp+δ (dx).[Hint: By Exercise 7.2.6, K ∈RHp(dx) is equivalent to the fact that K−1 ∈Ap′(K dx),

and the index p′ can be improved by Exercise 7.2.5 (d).]

7.2.8. (a) Show that for any w ∈ A1 and any cube Q in Rn and a > 1 we have

ess.infQ

w≤ an[w]A1ess.inf

aQw .

(b) Prove that there is a constant Cn such that for all locally integrable functions f

on Rn and all cubes Q in Rn we have

ess.infQ

M( f )≤Cn ess.inf3Q

M( f ) ,

and an analogous statement is valid for Mc.

[Hint: Part (a): Use (7.1.18). Part (b): Apply part (a) to M( f )

12 , which is an A1 weight

in view of Theorem 7.2.7.]

7.2.9. ([223]) For a weight w ∈ A1(Rn) define a quantity r = 1+ 1

2n+1[w]A1

. Show

that

Mc(wr)

1r ≤ 2 [w]A1

w a.e.

[Hint: Fix a cube Q and consider the family FQ of all cubes obtained by subdividing

Q into a mesh of (2n)m subcubes of side length 2−mℓ(Q) for all m = 1,2, . . . . Define

MdQ( f )(x) = supR∈FQ,R∋x |R|−1

∫R | f |dy. Using Corollary 2.1.21 obtain

∫

Q∩MdQ(w)>λ

w(x)dx≤ 2nλ |x ∈ Q : MdQ(w)(x)> λ|

for λ > wQ = 1|Q|

∫Q wdt. Multiply by λ δ−1 and integrate to obtain

∫

QMd

Q(w)δwdx≤ (wQ)

δ∫

Qwdx+

2nδ

δ +1

∫

QMd

Q(w)δ+1dx .

Replace w by wk = min(k,w) and select δ = 12n+1[w]A1

to deduce

1

|Q|

∫

Qwδ+1

k dx≤ 1

|Q|

∫

QMd

Q(wk)δwk dx≤ 2(wQ)

δ+1,

using [wk]A1≤ [w]A1

. Then let k→ ∞.]

7.2.10. Let 1 < p < ∞. Recall that a pair of weights (u,w) that satisfies

[u,w](Ap,Ap) = supQ cubes

in Rn

(1

|Q|

∫

Qudx

)(1

|Q|

∫

Qw− 1

p−1 dx

)p−1

< ∞

is said to be of class (Ap,Ap). The quantity [u,w](Ap,Ap) is called the (Ap,Ap) char-

acteristic constant of the pair (u,w).(a) Show that for any g ∈ L1

loc(Rn) with 0 < g <∞ a.e., the pair (g,M(g)) is of class

(Ap,Ap) with characteristic constant independent of f .

(b) If (u,w) is of class (Ap,Ap), then the Hardy–Littlewood maximal operator M

may not map Lp(w) to Lp(u).(c) Given g ∈ L1

loc(Rn) with 0 < g < ∞ a.e., conclude that Hardy–Littlewood max-

imal operator M maps Lp(M(g)dx) to Lp,∞(gdx) and also Lq(M(g)dx) to Lq(gdx)for any q with p < q < ∞.[Hint: Part (a): Use Holder’s inequality and Theorem 7.2.7. Part (b): Try the pair(M(g)1−p, |g|1−p

)for a suitable g. Part (c): Use Exercises 7.1.6 and 7.1.7.

]

7.3 The A∞ Condition 525

7.3 The A∞ Condition

In this section we examine more closely the class of all Ap weights. It turns out that

Ap weights possess properties that are p-independent but delicate enough to char-

acterize them without reference to a specific value of p. The Ap classes increase

as p increases, and it is only natural to consider their limit as p→ ∞. Not surpris-

ingly, a condition obtained as a limit of the Ap conditions as p→ ∞ provides some

unexpected but insightful characterizations of the class of all Ap weights.

7.3.1 The Class of A∞ Weights

Let us start by recalling a simple consequence of Jensen’s inequality:

(∫

X|h(t)|q dµ(t)

) 1q

≥ exp

(∫

Xlog |h(t)|dµ(t)

), (7.3.1)

which holds for all measurable functions h on a probability space (X ,µ) and all

0 < q < ∞. See Exercise 1.1.3(b). Moreover, part (c) of the same exercise says that

the limit of the expressions on the left in (7.3.1) as q→ 0 is equal to the expression

on the right in (7.3.1).

We apply (7.3.1) to the function h = w−1 for some weight w in Ap with q =1/(p−1). We obtain

w(Q)

|Q|

(1

|Q|

∫

Qw(t)−

1p−1 dt

)p−1

≥ w(Q)

|Q| exp

(1

|Q|

∫

Qlogw(t)−1 dt

), (7.3.2)

and the limit of the expressions on the left in (7.3.2) as p → ∞ is equal to the ex-

pression on the right in (7.3.2). This observation provides the motivation for the

following definition.

Definition 7.3.1. A weight w is called an A∞ weight if

[w]A∞ = supQ cubes in Rn

(1

|Q|

∫

Qw(t)dt

)exp

(1

|Q|

∫

Qlogw(t)−1 dt

)< ∞ .

The quantity [w]A∞ is called the A∞ characteristic constant of w.

It follows from the previous definition and (7.3.2) that for all 1≤ p < ∞ we have

[w]A∞ ≤ [w]Ap .

This means that ⋃

1≤p<∞

Ap ⊆ A∞ , (7.3.3)


but the remarkable thing is that equality actually holds in (7.3.3), a deep property

that requires some work.

Before we examine this and other characterizations of A∞ weights, we discuss

some of their elementary properties.

Proposition 7.3.2. Let w ∈ A∞. Then

(1) [δ λ (w)]A∞ = [w]A∞ , where δ λ (w)(x) = w(λx1, . . . ,λxn) and λ > 0.

(2) [τz(w)]A∞ = [w]A∞ , where τz(w)(x) = w(x− z), z ∈ Rn.

(3) [λw]A∞ = [w]A∞ for all λ > 0.

(4) [w]A∞ ≥ 1.

(5) The following is an equivalent characterization of the A∞ characteristic constant

of w:

[w]A∞ = supQcubesin Rn

suplog | f | ∈ L1(Q)∫

Q | f |wdt > 0

w(Q)∫

Q | f (t)|w(t)dtexp

(1

|Q|

∫

Qlog | f (t)|dt

).

(6) The measure w(x)dx is doubling; precisely, for all λ > 1 and all cubes Q we

have

w(λQ)≤ 2λn

[w]λn

A∞w(Q) .

As usual, λQ here denotes the cube with the same center as Q and side length

λ times that of Q.

We note that estimate (6) is not as good as λ → ∞ but it can be substantially

improved using the case λ = 2. We refer to Exercise 7.3.1 for an improvement.

Proof. Properties (1)–(3) are elementary, while property (4) is a consequence of

Exercise 1.1.3(b). To show (5), first observe that by taking f = w−1, the expression

on the right in (5) is at least as big as [w]A∞ . Conversely, (7.3.1) gives

exp

(1

|Q|

∫

Qlog

(| f (t)|w(t)

)dt

)≤ 1

|Q|

∫

Q| f (t)|w(t)dt ,

which, after a simple algebraic manipulation, can be written as

w(Q)∫Q | f |wdt

exp

(1

|Q|

∫

Qlog | f |dt

)≤ w(Q)

|Q| exp

(− 1

|Q|

∫

Qlog |w|dt

),

whenever f does not vanish almost everywhere on Q. Taking the supremum over all

such f and all cubes Q in Rn, we obtain that the expression on the right in (5) is at

most [w]A∞ .

To prove the doubling property for A∞ weights, we fix λ > 1 and we apply prop-

erty (5) to the cube λQ in place of Q and to the function

f =

c on Q,

1 on Rn \Q,(7.3.4)

where c is chosen so that c1/λ n= 2[w]A∞ . We obtain

w(λQ)

w(λQ\Q)+ cw(Q)exp

( logc

λ n

)≤ [w]A∞ ,

which implies (6) if we take into account the chosen value of c.

7.3.2 Characterizations of A∞ Weights

Having established some elementary properties of A∞ weights, we now turn to some

of their deeper properties, one of which is that every A∞ weight lies in some Ap

for p < ∞. It also turns out that A∞ weights are characterized by the reverse Holder

property, which as we saw is a fundamental property of Ap weights. The following

is the main theorem of this section.

Theorem 7.3.3. Suppose that w is a weight. Then w is in A∞ if and only if any one

of the following conditions holds:

(a) There exist 0 < γ ,δ < 1 such that for all cubes Q in Rn we have

∣∣x ∈ Q : w(x)≤ γ Avg Qw∣∣≤ δ |Q| .

(b) There exist 0 < α,β < 1 such that for all cubes Q and all measurable subsets A

of Q we have

|A| ≤ α |Q| =⇒ w(A)≤ β w(Q) .

(c) The reverse Holder condition holds for w, that is, there exist 0 <C1,ε < ∞ such

that for all cubes Q we have

(1

|Q|

∫

Qw(t)1+ε dt

) 11+ε

≤ C1

|Q|

∫

Qw(t)dt .

(d) There exist 0 <C2,ε0 < ∞ such that for all cubes Q and all measurable subsets

A of Q we have

w(A)

w(Q)≤C2

( |A||Q|

)ε0

.

(e) There exist 0 < α ′,β ′ < 1 such that for all cubes Q and all measurable subsets

A of Q we have

w(A)< α ′w(Q) =⇒ |A|< β ′ |Q| .

(f) There exist p,C3 < ∞ such that [w]Ap ≤C3. In other words, w lies in Ap for some

p ∈ [1,∞).All the constants C1,C2,C3,α,β ,γ ,δ ,α

′,β ′,ε ,ε0, and p in (a)–(f) depend only

on the dimension n and on [w]A∞ . Moreover, if any of the statements in (a)–(f) is

valid, then so is any other statement in (a)–(f) with constants that depend only on

the dimension n and the constants that appear in the assumed statement.

Proof. The proof follows from the sequence of implications

w ∈ A∞ =⇒ (a) =⇒ (b) =⇒ (c) =⇒ (d) =⇒ (e) =⇒ ( f ) =⇒ w ∈ A∞ .

At each step we keep track of the way the constants depend on the constants of the

previous step. This is needed to validate the last assertion of the theorem.

w ∈ A∞ =⇒ (a)Fix a cube Q. Since multiplication of an A∞ weight with a positive scalar does

not alter its A∞ characteristic, we may assume that∫

Q logw(t)dt = 0. This implies

that AvgQ w≤ [w]A∞ . Then we have

∣∣x ∈ Q : w(x)≤ γAvgQ

w∣∣ ≤

∣∣x ∈ Q : w(x)≤ γ [w]A∞∣∣

=∣∣x ∈ Q : log(1+w(x)−1)≥ log(1+(γ [w]A∞)

−1)∣∣

≤ 1

log(1+(γ [w]A∞)−1)

∫

Qlog

1+w(t)

w(t)dt

=1

log(1+(γ [w]A∞)−1)

∫

Qlog(1+w(t))dt

≤ 1

log(1+(γ [w]A∞)−1)

∫

Qw(t)dt

≤ [w]A∞ |Q|log(1+(γ [w]A∞)

−1)

=1

2|Q| ,

which proves (a ) with γ = [w]−1A∞(e2[w]A∞ −1)−1 and δ = 1

2.

(a) =⇒ (b)Let Q be fixed and let A be a subset of Q with w(A) > βw(Q) for some β to be

chosen later. Setting S = Q\A, we have w(S)< (1−β )w(Q). We write S = S1∪S2,

where

S1 = x ∈ S : w(x)> γAvg Qw and S2 = x ∈ S : w(x)≤ γAvg Qw .

For S2 we have |S2| ≤ δ |Q| by assumption (a ). For S1 we use Chebyshev’s inequality

to obtain

|S1| ≤1

γAvgQ

w

∫

Sw(t)dt =

|Q|γ

w(S)

w(Q)≤ 1−β

γ|Q| .

Adding the estimates for |S1| and |S2|, we obtain

|S| ≤ |S1|+ |S2| ≤1−βγ|Q|+δ |Q|=

(δ +

1−βγ

)|Q| .

Choosing numbers α,β in (0,1) such that δ + 1−βγ = 1−α , for example α = 1−δ

2

and β = 1− (1−δ )γ2

, we obtain |S| ≤ (1−α)|Q|, that is, |A|> α|Q|.(b) =⇒ (c)

This was proved in Corollary 7.2.4. To keep track of the constants, we note that

the choices

ε =− 1

2logβ

log2n− logαand C1 = 1+

(2nα−1)ε

1− (2nα−1)εβ

as given in (7.2.6) and (7.2.7) serve our purposes.

(c) =⇒ (d )We apply first Holder’s inequality with exponents 1+ ε and (1+ ε)/ε and then

the reverse Holder estimate to obtain

∫

Aw(x)dx ≤

(∫

Aw(x)1+ε dx

) 11+ε

|A| ε1+ε

≤(

1

|Q|

∫

Qw(x)1+ε dx

) 11+ε

|Q| 11+ε |A| ε

1+ε

≤ C1

|Q|

∫

Qw(x)dx |Q| 1

1+ε |A| ε1+ε ,

which givesw(A)

w(Q)≤C1

( |A||Q|

) ε1+ε

.

This proves (d ) with ε0 =ε

1+ε and C2 =C1.

(d ) =⇒ (e)Pick an 0 < α ′′ < 1 small enough that β ′′ = C2(α

′′)ε0 < 1. It follows from (d )

that

|A|< α ′′|Q| =⇒ w(A)< β ′′w(Q) (7.3.5)

for all cubes Q and all A measurable subsets of Q. Replacing A by Q\A, the impli-

cation in (7.3.5) can be equivalently written as

|A| ≥ (1−α ′′)|Q| =⇒ w(A)≥ (1−β ′′)w(Q) .

In other words, for measurable subsets A of Q we have

w(A)< (1−β ′′)w(Q) =⇒ |A|< (1−α ′′)|Q| , (7.3.6)

which is the statement in (e ) if we set α ′ = (1− β ′′) and β ′ = 1−α ′′. Note that

(7.3.5) and (7.3.6) are indeed equivalent.

(e) =⇒ ( f )We begin by examining condition (e ), which can be written as

∫

Aw(t)dt ≤ α ′

∫

Qw(t)dt =⇒

∫

Aw(t)−1w(t)dt ≤ β ′

∫

Qw(t)−1w(t)dt ,

or, equivalently, as

µ(A)≤ α ′µ(Q) =⇒∫

Aw(t)−1 dµ(t)≤ β ′

∫

Qw(t)−1 dµ(t)

after defining the measure dµ(t) = w(t)dt. As we have already seen, the assertions

in (7.3.5) and (7.3.6) are equivalent. Therefore, we may use Exercise 7.3.2 to deduce

that the measure µ is doubling, i.e., it satisfies property (7.2.9) for some constant

Cn =Cn(α′,β ′), and hence the hypotheses of Corollary 7.2.4 are satisfied. We con-

clude that the weight w−1 satisfies a reverse Holder estimate with respect to the

measure µ , that is, if γ ,C are defined as in (7.2.11) and (7.2.12) [in which α is re-

placed by α ′, β by β ′, and Cn is the doubling constant of w(x)dx], then we have

(1

µ(Q)

∫

Qw(t)−1−γ dµ(t)

) 11+γ

≤ C

µ(Q)

∫

Qw(t)−1 dµ(t) (7.3.7)

for all cubes Q in Rn. Setting p = 1+ 1γ and raising to the pth power, we can rewrite

(7.3.7) as the Ap condition for w. We can therefore take C3 = Cp to conclude the

proof of (f).

( f ) =⇒ w ∈ A∞This is trivial, since [w]A∞ ≤ [w]Ap .

An immediate consequence of the preceding theorem is the following result

relating A∞ to Ap.

Corollary 7.3.4. The following equality is valid:

A∞ =⋃

1≤p<∞

Ap.

Exercises

7.3.1. Let λ > 0, Q be a cube in Rn, and w ∈ A∞(Rn).

(a) Show that property (6) in Proposition 7.3.2 can be improved to

w(λQ)≤minε>0

(1+ ε)λn[w]λ

n

A∞−1

εw(Q) .

(b) Prove that

w(λQ)≤ (2λ )2n(1+log2[w]A∞ )w(Q) .[Hint: Part (a): Take c in (7.3.4) such that c1/λ n

= (1+ ε)[w]A∞ . Part (b): Use the

estimate in property (6) of Proposition 7.3.2 with λ = 2.]

7.3.2. Suppose that µ is a positive Borel measure on Rn with the property that for

all cubes Q and all measurable subsets A of Q we have

|A|< α|Q| =⇒ µ(A)< βµ(Q)

for some fixed 0 < α,β < 1. Show that µ is doubling [i.e., it satisfies (7.2.9)].[Hint: Use that |S|> (1−α)|Q| ⇒ µ(S)> (1−β )µ(Q) when S Q.

]

7.3.3. Prove that a weight w is in Ap if and only if both w and w− 1

p−1 are in A∞.[Hint: You may want to use the result of Exercise 7.2.2.

]

7.3.4. ([33], [343]) Prove that if P(x) is a polynomial of degree k in Rn, then

log |P(x)| is in BMO with norm depending only on k and n and not on the coef-

ficients of the polynomial.[Hint: Use that all norms on the finite-dimensional space of polynomials of degree

at most k are equivalent to show that |P(x)| satisfies a reverse Holder inequality.

Therefore, |P(x)| is an A∞ weight and thus Exercise 7.2.3 (c) is applicable.]

7.3.5. Show that the product of two A1 weights may not be an A∞ weight.

7.3.6. Let g be in Lp(w) for some 1≤ p≤ ∞ and w ∈ Ap. Prove that g ∈ L1loc(R

n).[Hint: Let B be a ball. In the case p < ∞, write

∫B |g|dx =

∫B(|g|w

− 1p )w

1p dx and

apply Holder’s inequality. In the case p = ∞, use that w ∈ Ap0for some p0 < ∞.

]

7.3.7. ([278]) Show that a weight w lies in A∞ if and only if there exist γ ,C > 0 such

that for all cubes Q we have

w(

x ∈ Q : w(x)> λ)≤Cλ

∣∣x ∈ Q : w(x)> γλ∣∣

for all λ > AvgQ w.[Hint: The displayed condition easily implies that

1

|Q|

∫

Qw1+ε

k dx≤(w(Q)

|Q|)ε+1

+C′δγ1+ε

1

|Q|

∫

Qw1+ε

k dx ,

where k > 0, wk = min(w,k) and δ = ε/(1+ ε). Take ε > 0 small enough to obtain

the reverse Holder condition (c ) in Theorem 7.3.3 for wk. Let k → ∞ to obtain the

same conclusion for w. Conversely, find constants γ ,δ ∈ (0,1) as in condition (a) of

Theorem 7.3.3 and for λ > AvgQ w write the set w > λ∩Q as a union of maximal

dyadic cubes Q j such that λ < AvgQ jw ≤ 2nλ for all j. Then w(Q j) ≤ 2nλ |Q j| ≤

2nλ1−δ |Q j ∩w > γλ| and the required conclusion follows by summing on j.

]

7.4 Weighted Norm Inequalities for Singular Integrals

We now address a topic of great interest in the theory of singular integrals, their

boundedness properties on weighted Lp spaces. It turns out that a certain amount

of regularity must be imposed on the kernels of these operators to obtain the afore-

mentioned weighted estimates.

7.4.1 Singular Integrals of Non Convolution type

We introduce some definitions.

Definition 7.4.1. Let 0 < δ ,A < ∞. A function K(x,y) defined for x,y ∈ Rn with

x = y is called a standard kernel (with constants δ and A) if

|K(x,y)| ≤ A

|x− y|n , x = y, (7.4.1)

and whenever |x− x′| ≤ 12

max(|x− y|, |x′− y|

)we have

|K(x,y)−K(x′,y)| ≤ A|x− x′|δ(|x− y|+ |x′− y|)n+δ

(7.4.2)

and also when |y− y′| ≤ 12

max(|x− y|, |x− y′|

)we have

|K(x,y)−K(x,y′)| ≤ A|y− y′|δ(|x− y|+ |x− y′|)n+δ

. (7.4.3)

The class of all kernels that satisfy (7.4.1), (7.4.2), and (7.4.3) is denoted by

SK(δ ,A).

Definition 7.4.2. Let 0 < δ ,A <∞ and K in SK(δ ,A). A Calderon–Zygmund opera-

tor associated with K is a linear operator T defined on S (Rn) that admits a bounded

extension on L2(Rn), ∥∥T ( f )∥∥

L2 ≤ B∥∥ f

∥∥L2 , (7.4.4)

and that satisfies

T ( f )(x) =∫

RnK(x,y) f (y)dy (7.4.5)

for all f ∈ C ∞0 and x not in the support of f . The class of all Calderon–Zygmund

operators associated with kernels in SK(δ ,A) that are bounded on L2 with norm at

most B is denoted by CZO(δ ,A,B). Note that there is no unique T associated with

a given K. Given a Calderon–Zygmund operator T in CZO(δ ,A,B), we define the

truncated operator T (ε) as

T (ε)( f )(x) =∫

|x−y|>εK(x,y) f (y)dy

7.4 Weighted Norm Inequalities for Singular Integrals 533

and the maximal operator associated with T as follows:

T (∗)( f )(x) = supε>0

∣∣T (ε)( f )(x)∣∣ .

We note that if T is in CZO(δ ,A,B), then T (ε)( f ) and T (∗)( f ) are well defined

for all f in⋃

1≤p<∞Lp(Rn). It is also well defined whenever f is locally integrable

and satisfies∫|x−y|≥ε | f (y)| |x− y|−ndy < ∞ for all x ∈ Rn and ε > 0.

The class of kernels in SK(δ ,A) extends the family of convolution kernels that

satisfy conditions (5.3.10), (5.3.11), and (5.3.12). Obviously, the associated operators

in CZO(δ ,A,B) generalize the associated convolution operators.

A fundamental property of operators in CZO(δ ,A,B) is that they have bounded

extensions on all the Lp(Rn) spaces and also from L1(Rn) to weak L1(Rn). This is

proved via an adaptation of Theorem 5.3.3; see Theorem 4.2.2 in [131]. There are

analogous results for the maximal counterparts T (∗) of elements of CZO(δ ,A,B).In fact, an analogue of Theorem 5.3.5 yields that T (∗) is Lp bounded for 1 < p < ∞and weak type (1,1); this result is contained in Theorem 4.2.4 in [131].

We discuss weighted inequalities for singular integrals for general operators in

CZO(δ ,A,B). In Subsections 7.4.2 and 7.4.3, the reader may wish to replace kernels

in SK(δ ,A) by the more familiar functions K(x) defined on Rn \ 0 that satisfy

(5.3.10), (5.3.11), and (5.3.12).

7.4.2 A Good Lambda Estimate for Singular Integrals

The following theorem is the main result of this section.

Theorem 7.4.3. Let 1 ≤ p ≤ ∞, w ∈ Ap, and T in CZO(δ ,A,B). Then there exist

positive constants1 C0 =C0(n, p, [w]Ap), ε0 = ε0(n, p, [w]Ap), and c0(n,δ ), such that

if γ0 = c0(n,δ )/A, then for all 0 < γ < γ0 we have

w(T (∗)( f )> 3λ∩M( f )≤ γλ

)≤C0γ

ε0(A+B)ε0w(T (∗)( f )> λ

), (7.4.6)

for all locally integrable functions f for which

∫

|x−y|≥ε| f (y)| |x− y|−ndy < ∞

for all x ∈ Rn and ε > 0. Here M denotes the Hardy–Littlewood maximal operator.

Proof. We write the open set

Ω = T (∗)( f )> λ=⋃

j

Q j ,

1 the dependence on p is relevant only when p < ∞

where Q j are the Whitney cubes (see Appendix J). We set

Q∗j = 10√

nQ j ,

Q∗∗j = 10√

nQ∗j ,

where aQ denotes the cube with the same center as Q whose side length is aℓ(Q),where ℓ(Q) is the side length of Q. We note that in view of the properties of the

Whitney cubes, the distance from Q j to Ω c is at most 4√

nℓ(Q j). But the distance

from Q j to the boundary of Q∗j is (5√

n− 12)ℓ(Q j), which is bigger than 4

√nℓ(Q j).

Therefore, Q∗j must meet Ω c and for every cube Q j we fix a point y j in Ω c∩Q∗j . See

Figure 7.1.

Qj

Qj

Qj

**

* yj

.

..

.

xzj

t

c

(50 -

-

n 5 n ) l (Q )j

(5 n ) l (Q )j

12

ΩΩ

Fig. 7.1 A picture of the proof.

We also fix f in⋃

1≤p<∞Lp(Rn), and for each j we write f = fj

0 + fj∞, where

fj

0 = f χQ∗∗j is the part of f near Q j and fj∞ = f χ(Q∗∗j )c is the part of f away from Q j.

We now claim that the following estimate is true:

∣∣Q j ∩T (∗)( f )> 3λ∩M( f )≤ γλ∣∣≤Cn γ (A+B)

∣∣Q j

∣∣ . (7.4.7)

Once the validity of (7.4.7) is established, we apply Theorem 7.3.3 (d) when p = ∞or Proposition 7.2.8 when p < ∞ to obtain constants ε0,C2 > 0, which depend on

[w]Ap , p, n when p < ∞ and on [w]A∞ and n when p = ∞, such that

w(Q j ∩T (∗)( f )> 3λ∩M( f )≤ γλ

)≤C2 (Cn)

ε0 γε0 (A+B)ε0 w(Q j) .

Then a simple summation in j gives (7.4.6) with C0 = C2(Cn)ε0 , and recall that C2

and ε0 depend on n and [w]Ap and on p if p < ∞.

In proving estimate (7.4.7), we may assume that for each cube Q j there exists a

z j ∈ Q j such that M( f )(z j)≤ γ λ ; otherwise, the set on the left in (7.4.7) is empty.

We now invoke Theorem 4.2.4 in [131], which states that T (∗) maps L1(Rn) to

L1,∞(Rn) with norm at most C(n)(A+B). We have the estimate

∣∣Q j ∩T (∗)( f )> 3λ∩M( f )≤ γλ∣∣≤ Iλ0 + Iλ∞ , (7.4.8)

where

Iλ0 =∣∣Q j ∩T (∗)( f

j0 )> λ∩M( f )≤ γλ

∣∣ ,Iλ∞ =

∣∣Q j ∩T (∗)( f j∞)> 2λ∩M( f )≤ γλ

∣∣ .

To control Iλ0 we note that fj

0 is in L1(Rn) and we argue as follows:

Iλ0 ≤∣∣T (∗)( f

j0 )> λ

∣∣

≤ ‖T (∗)‖L1→L1,∞

λ

∫

Rn| f j

0 (x)|dx

≤ C(n)(A+B)|Q∗∗j |λ

1

|Q∗∗j |

∫

Q∗∗j| f (x)|dx

≤ C(n)(A+B)|Q∗∗j |λ

Mc( f )(z j)

≤ C(n)(A+B)|Q∗∗j |λ

M( f )(z j)

≤ C(n)(A+B)|Q∗∗j |λ

λ γ

= Cn (A+B)γ |Q j| .

(7.4.9)

Next we claim that Iλ∞ = 0 if we take γ sufficiently small. We first show that for all

x ∈ Q j we have

supε>0

∣∣T (ε)( f j∞)(x)−T (ε)( f j

∞)(y j)∣∣≤C

(1)n,δ AM( f )(z j) . (7.4.10)

Indeed, let us fix an ε > 0. We have

∣∣T (ε)( f j∞)(x)−T (ε)( f j

∞)(y j)∣∣ =

∣∣∣∣∫

|t−x|>ε

K(x, t) f j∞(t)dt−

∫

|t−y j|>ε

K(y j, t) f j∞(t)dt

∣∣∣∣

≤ L1 +L2 +L3 ,


where

L1 =

∣∣∣∣∫

|t−y j |>ε

[K(x, t)−K(y j, t)

]f j∞(t)dt

∣∣∣∣ ,

L2 =

∣∣∣∣∫

|t−x|>ε|t−y j |≤ε

K(x, t) f j∞(t)dt

∣∣∣∣ ,

L3 =

∣∣∣∣∫

|t−x|≤ε|t−y j |>ε

K(x, t) f j∞(t)dt

∣∣∣∣ ,

in view of identity (5.4.7).

We now make a couple of observations. For t /∈ Q∗∗j , x,z j ∈ Q j, and y j ∈ Q∗j we

have

3

4≤ |t− x||t− y j|

≤ 5

4,

48

49≤ |t− x||t− z j|

≤ 50

49. (7.4.11)

Indeed,

|t− y j| ≥ (50n−5√

n)ℓ(Q j)≥ 44nℓ(Q j)

and

|x− y j| ≤1

2

√nℓ(Q j)+

√n10

√nℓ(Q j)≤ 11nℓ(Q j)≤

1

4|t− y j| .

Using this estimate and the inequalities

3

4|t− y j| ≤ |t− y j|− |x− y j| ≤ |t− x| ≤ |t− y j|+ |x− y j| ≤

5

4|t− y j| ,

we obtain the first estimate in (7.4.11). Likewise, we have

|x− z j| ≤√

nℓ(Q j)≤ nℓ(Q j)

and

|t− z j| ≥ (50n− 12)ℓ(Q j)≥ 49nℓ(Q j) ,

and these give

48

49|t− z j| ≤ |t− z j|− |x− z j| ≤ |t− x| ≤ |t− z j|+ |x− z j| ≤

50

49|t− z j| ,

yielding the second estimate in (7.4.11).

Since |x− y j| ≤ 12|t− y j| ≤ 1

2max

(|t− x|, |t− y j|

), we have

|K(x, t)−K(y j, t)| ≤A|x− y j|δ

(|t− x|+ |t− y j|)n+δ≤C′n,δA

ℓ(Q j)δ

|t− z j|n+δ;


hence, we obtain

L1 ≤∫

|t−z j |≥49nℓ(Q j)

C′n,δAℓ(Q j)

δ

|t− z j|n+δ| f (t)|dt ≤C′′n,δAM( f )(z j)

using Theorem 2.1.10. Using (7.4.11) we deduce

L2 ≤∫

|t−z j |≤ 54 · 49

48 ε

A

|x− t|n χ|t−x|≥ε | f j∞(t)|dt ≤C′nAM( f )(z j) .

Again using (7.4.11), we obtain

L3 ≤∫

|t−z j |≤ 4948 ε

A

|x− t|n χ|t−x|≥ 34 ε| f j∞(t)|dt ≤C′′n AM( f )(z j) .

This proves (7.4.10) with constant C(1)n,δ =C′′

n,δ +C′n +C′′n .

Having established (7.4.10), we next claim that

supε>0

∣∣T (ε)( f j∞)(y j)

∣∣≤ T (∗)( f )(y j)+C(2)n AM( f )(z j) . (7.4.12)

To prove (7.4.12) we fix a cube Q j and ε > 0. We let R j be the smallest number such

that

Q∗∗j ⊆ B(y j,R j) .

See Figure 7.2. We consider the following two cases.

Qj

Qj

Qj

**

* yj.

Rj

Fig. 7.2 The ball B(y j,R j).

Case (1): ε ≥ R j. Since Q∗∗j ⊆ B(y j,ε), we have B(y j,ε)c ⊆ (Q∗∗j )c and therefore

T (ε)( f j∞)(y j) = T (ε)( f )(y j) ,

so (7.4.12) holds easily in this case.

Case (2): 0 < ε < R j. Note that if t ∈ (Q∗∗j )c, then |t − y j| ≥ 40nℓ(Q j). On the

other hand, R j ≤ diam(Q∗∗j ) = 100n32 ℓ(Q j). This implies that

R j ≤ 5√

n

2|t− y j| , when t ∈ (Q∗∗j )c .

Notice also that in this case we have B(y j,R j)c ⊆ (Q∗∗j )c, hence

T (R j)( f j∞)(y j) = T (R j)( f )(y j) .

Therefore, we have

∣∣T (ε)( f j∞)(y j)

∣∣ ≤∣∣T (ε)( f j

∞)(y j)−T (R j)( f j∞)(y j)

∣∣+∣∣T (R j)( f )(y j)

∣∣

≤∫

ε≤|y j−t|≤R j

|K(y j, t)| | f j∞(t)|dt +T (∗)( f )(y j)

≤∫

25√

nR j≤|y j−t|≤R j

|K(y j, t)|| f j∞(t)|dt +T (∗)( f )(y j)

≤A( 2

5√

n)−n

Rnj

∫

|z j−t|≤ 54 · 49

48 R j

| f (t)|dt +T (∗)( f )(y j)

≤ C(2)n AM( f )(z j)+T (∗)( f )(y j) ,

where in the penultimate estimate we used (7.4.11). The proof of (7.4.12) follows

with the required bound C(2)n A.

Combining (7.4.10) and (7.4.12), we obtain

T (∗)( f j∞)(x)≤ T (∗)( f )(y j)+

(C(1)n,δ +C

(2)n

)AM( f )(z j) .

Recalling that y j /∈Ω and that M( f )(z j)≤ γλ , we deduce

T (∗)( f j∞)(x)≤ λ +

(C(1)n,δ +C

(2)n

)Aγλ .

Setting γ0 =(C(1)n,δ +C

(2)n

)−1A−1 = c0(n,δ )A

−1, for 0 < γ < γ0, we have that the set

Q j ∩T (∗)( f j∞)> 2λ∩M( f )≤ γλ

is empty. This shows that the quantity Iγ∞ vanishes if γ is smaller than γ0. Returning

to (7.4.8) and using the estimate (7.4.9) proved earlier, we conclude the proof of

(7.4.7), which, as indicated earlier, implies the theorem.

Remark 7.4.4. We observe that for any δ > 0, estimate (7.4.6) also holds for the

operator

T(∗)δ ( f )(x) = sup

ε≥δ|T (ε)( f )(x)| (7.4.13)

with the same constant (which is independent of δ ).

To see the validity of (7.4.6) for T(∗)δ , it suffices to prove

∣∣T (∗)δ ( f j

∞)(y j)∣∣≤ T

(∗)δ ( f )(y j)+C

(2)n AM( f )(z j) , (7.4.14)

which is a version of (7.4.12) with T (∗) replaced by T(∗)δ . The following cases arise:

Case (1′): R j ≤ δ ≤ ε or δ ≤ R j ≤ ε . Here, as in Case (1) we have

|T (ε)( f j∞)(y j)|= |T (ε)( f )(y j)| ≤ T

(∗)δ ( f )(y j) .

Case (2′): δ ≤ ε < R j. As in Case (2) we have

T (R j)( f j∞)(y j) = T (R j)( f )(y j),

thus

∣∣T (ε)( f j∞)(y j)

∣∣ ≤∣∣T (ε)( f j

∞)(y j)−T (R j)( f j∞)(y j)

∣∣+∣∣T (R j)( f )(y j)

∣∣ .

As in the proof of Case (2), we bound the first term on the right of the last displayed

expression by C(2)n AM( f )(z j) while the second term is at most T

(∗)δ ( f )(y j).

7.4.3 Consequences of the Good Lambda Estimate

Having obtained the important good lambda weighted estimate for singular inte-

grals, we now pass to some of its consequences. We begin with the following lemma:

Lemma 7.4.5. Let 1≤ p < ∞, ε > 0, w ∈ Ap, x ∈Rn, and f ∈ Lp(w). Then we have

∫

|x−y|≥ε

| f (y)||x− y|n dy≤C00(w,n, p,x,ε)

∥∥ f∥∥

Lp(w)

for some constant C00 depending on the stated parameters. In particular, T (ε)( f )and T (∗)( f ) are defined for f ∈ Lp(w).

Proof. For each ε > 0 and x pick a cube Q0 = Q0(x,ε) of side length cnε (for some

constant cn) such that Q0 B(x,ε). Set Q j = 2 jQ0 for j ≥ 0. We have

∫

|y−x|≥ε

| f (y)||x− y|n dy ≤ Cn

∞

∑j=0

(2 jε)−n

∫

Q j+1\Q j

| f (y)|dy

≤ Cn

∞

∑j=1

(1

|Q j|

∫

Q j

| f (y)|pwdy

) 1p(

1

|Q j|

∫

Q j

w− p′

p dy

) 1p′

≤ Cn [w]1p

Ap

∞

∑j=1

(∫

Q j

| f (y)|pwdy

) 1p(

1

w(Q j)

) 1p

≤ Cn [w]1p

Ap

∥∥ f∥∥

Lp(w)

∞

∑j=1

(w(Q j)

)− 1p .

But Proposition 7.2.8 gives for some δ = δ (n, p, [w]Ap) that

w(Q0)

w(Q j)≤C(n, p, [w]Ap)

|Q0|δ|Q j|δ

,


w(Q j)− 1

p ≤C′(n, p, [w]Ap)2− j nδ

p w(Q0)− 1

p .

In view of this estimate, the previous series converges. Note that C′ and thus C00

depend on [w]Ap ,n, p,x,ε , and w(Q0).This argument is also valid in the case p = 1 by an obvious modification.

Theorem 7.4.6. Let A,B,β > 0 and let T be a CZO(β ,A,B). Then given 1 < p <∞,

there is a constant Cp =Cp(n,β , [w]Ap) such that

∥∥T (∗)( f )∥∥

Lp(w)≤Cp (A+B)

∥∥ f∥∥

Lp(w)(7.4.15)

for all w∈ Ap and f ∈ Lp(w). There is also a constant C1 =C1(n,β , [w]A1) such that

∥∥T (∗)( f )∥∥

L1,∞(w)≤C1 (A+B)

∥∥ f∥∥

L1(w)(7.4.16)

for all w ∈ A1 and f ∈ L1(w).

Proof. This theorem is a consequence of the estimate proved in the previous the-

orem. For technical reasons, it is useful to fix a δ > 0 and work with the auxil-

iary maximal operator T(∗)δ defined in (7.4.13) instead of T (∗). We begin by taking

1 λ)

dλ

= 3p

∫ ∞

0pλ p−1w

(T (∗)

δ ( f )> 3λ)

dλ ,

which we control by

3p

∫ ∞

0pλ p−1w

(T (∗)

δ ( f )> 3λ∩M( f )≤ γλ)

dλ

+ 3p

∫ ∞

0pλ p−1w

(M( f )> γλ

)dλ .

Using Theorem 7.4.3 (or rather Remark 7.4.4), there are C0 = C0(n, [w]Ap), ε0 =

ε0(n, [w]Ap), and γ0 = c0(n,β )A−1, such that the preceding displayed expression is

bounded by

3pC0γε0(A+B)ε0

∫ ∞

0pλ p−1w

(T (∗)

δ ( f )> λ)

dλ

+3p

γ p

∫ ∞

0pλ p−1w

(M( f )> λ

)dλ ,

which is equal to

3pC0γε0(A+B)ε0

∥∥T(∗)δ ( f )

∥∥p

Lp(w)+

3p

γ p

∥∥M( f )∥∥p

Lp(w).

Taking γ = min(

12c0(n,β )A

−1, 12(2C03p)

− 1ε0 (A+B)−1

)< γ0, we conclude that

∥∥T(∗)δ ( f )

∥∥p

Lp(w)

≤ 1

2

∥∥T(∗)δ ( f )

∥∥p

Lp(w)+Cp(n,β , [w]Ap)(A+B)p

∥∥M( f )∥∥p

Lp(w).

(7.4.17)

We now prove a similar estimate when p = 1. For f ∈ L1(w) and w ∈ A1 we have

3λw(

T(∗)δ ( f )> 3λ

)

≤ 3λw(

T(∗)δ ( f )> 3λ

∩M( f )≤ γλ

)+3λw

(M( f )> γλ

),

and this expression is controlled by

3λC0γε0(A+B)ε0w

(T(∗)δ ( f )> λ

)+

3

γ

∥∥M( f )∥∥

L1,∞(w).

Recalling that γ0 = c0(n,β )A−1 and choosing γ = min

(12γ0,

12(6C0)

− 1ε0 (A+B)−1

),

it follows that

∥∥T(∗)δ ( f )

∥∥L1,∞(w)

≤ 1

2

∥∥T(∗)δ ( f )

∥∥L1,∞(w)

+C1(n,β , [w]A1)(A+B)

∥∥M( f )∥∥

L1,∞(w).

(7.4.18)

Estimate (7.4.15) would follow from (7.4.17) if we knew that ‖T (∗)δ ( f )‖Lp(w) < ∞

whenever 1 < p < ∞, w ∈ Ap and f ∈ Lp(w), while (7.4.16) would follow from

(7.4.18) if we had ‖T (∗)δ ( f )‖L1,∞(w) < ∞ whenever w ∈ A1 and f ∈ L1(w). Since we

do not know that these quantities are finite, a certain amount of work is needed.

To deal with this problem we momentarily restrict attention to a special class

of functions on Rn, the class of bounded functions with compact support. Such

functions are dense in Lp(w) when w ∈ Ap and 1 ≤ p < ∞; see Exercise 7.4.1. Let

h be a bounded function with compact support on Rn. Then T(∗)δ (h)≤C1δ

−n‖h‖L1

and T(∗)δ (h)(x)≤C2(h)|x|−n for x away from the support of h. It follows that

T(∗)δ (h)(x)≤C3(h,δ )(1+ |x|)−n

for all x ∈ Rn. Furthermore, if h is nonzero, then

M(h)(x)≥ C4(h)

(1+ |x|)n,

and therefore for w ∈ A1,

∥∥T(∗)δ (h)

∥∥L1,∞(wdx)

≤C5(h,δ )∥∥M(h)

∥∥L1,∞(wdx)

< ∞ ,

while for 1 < p < ∞ and w ∈ Ap,

∫

Rn(T

(∗)δ (h)(x))pw(x)dx≤C5(h, p,δ )

∫

RnM(h)(x)pw(x)dx < ∞

in view of Theorem 7.1.9. Using these facts, (7.4.17), (7.4.18), and Theorem 7.1.9

once more, we conclude that for all δ > 0 and 1 < p < ∞ we have

∥∥T(∗)δ (h)

∥∥p

Lp(w)≤2Cp

∥∥M(h)∥∥p

Lp(w)≤ C′p[w]

pp−1

Ap

∥∥h∥∥p

Lp(w)=Cp

p

∥∥h∥∥p

Lp(w),

∥∥T(∗)δ (h)

∥∥L1,∞(w)

≤2C1

∥∥M(h)∥∥

L1,∞(w)≤ C1[w]A1

∥∥h∥∥

L1(w)=C1

∥∥h∥∥

L1(w),

(7.4.19)

whenever h a bounded function with compact support. The constants Cp, C′p, and Cp

depend only on the parameters n, β , p, and [w]Ap .

We now extend estimates (7.4.16) and (7.4.15) to functions in Lp(Rn,wdx). Given

1≤ p < ∞, w ∈ Ap, and f ∈ Lp(w), let

fN(x) = f (x)χ| f |≤Nχ|x|≤N .

Then fN is a bounded function with compact support that converges to f in Lp(w)(i.e., ‖ fN − f‖Lp(w) → 0 as N → ∞) by the Lebesgue dominated convergence theo-

rem. Also | fN | ≤ | f | for all N. Sublinearity and Lemma 7.4.5 give for all x ∈ Rn,

|T (∗)δ ( fN)(x)−T

(∗)δ ( f )(x)| ≤ T

(∗)δ ( f − fN)(x)

≤ AC00(w,n, p,x,δ )∥∥ fN− f

∥∥Lp(w)

,

and this converges to zero as N → ∞ since C00(w,n, p,x,δ )< ∞. Therefore

T(∗)δ ( f ) = lim

N→∞T(∗)δ ( fN)

pointwise, and Fatou’s lemma for weak type spaces [see Exercise 1.1.12 (d)] gives

for w ∈ A1 and f ∈ L1(w),

∥∥T(∗)δ ( f )

∥∥L1,∞(w)

=∥∥ liminf

N→∞T(∗)δ ( fN)

∥∥L1,∞(w)

≤ liminfN→∞

∥∥T(∗)δ ( fN)

∥∥L1,∞(w)

≤ C1 liminfN→∞

∥∥M( fN)∥∥

L1,∞(w)

≤ C1

∥∥M( f )∥∥

L1,∞(w),

since | fN | ≤ | f | for all N. An analogous argument gives the estimate

∥∥T(∗)δ ( f )

∥∥Lp(w)

≤Cp

∥∥ f∥∥

Lp(w)

for w ∈ Ap and f ∈ Lp(w) when 1 < p < ∞.

It remains to prove (7.4.15) and (7.4.16) for T (∗). But this is also an easy conse-

quence of Fatou’s lemma, since the constants Cp and C1 are independent of δ and

limδ→0

T(∗)δ ( f ) = T (∗)( f )

for all f ∈ Lp(w).

We end this subsection by making the comment that if a given T in CZO(δ ,A,B)is pointwise controlled by T (∗), then the estimates of Theorem 7.4.6 also hold for it.

This is the case for the Hilbert transform, the Riesz transforms, and other classical

singular integral operators.

7.4.4 Necessity of the Ap Condition

We have established the main theorems relating Calderon–Zygmund operators and

Ap weights, namely that such operators are bounded on Lp(w) whenever w lies in

Ap. It is natural to ask whether the Ap condition is necessary for the boundedness of

singular integrals on Lp. We end this section by indicating the necessity of the Ap

condition for the boundedness of the Riesz transforms on weighted Lp spaces.

Theorem 7.4.7. Let w be a weight in Rn and let 1≤ p <∞. Suppose that each of the

Riesz transforms R j is of weak type (p, p) with respect to w. Then w must be an Ap

weight. Similarly, let w be a weight in R. If the Hilbert transform H is of weak type

(p, p) with respect to w, then w must be an Ap weight.

Proof. We prove the n-dimensional case, n≥ 2. The one-dimensional case is essen-

tially contained in following argument, suitably adjusted.

Let Q be a cube and let f be a nonnegative function on Rn supported in Q that

satisfies AvgQ f > 0. Let Q′ be the cube that shares a corner with Q, has the same

length as Q, and satisfies x j ≥ y j for all 1≤ j ≤ n whenever x ∈Q′ and y ∈Q. Then

for x ∈ Q′ we have

∣∣∣∣n

∑j=1

R j( f )(x)

∣∣∣∣=Γ ( n+1

2)

πn+1

2

n

∑j=1

∫

Q

x j− y j

|x− y|n+1f (y)dy≥ Γ ( n+1

2)

πn+1

2

∫

Q

f (y)

|x− y|n dy .

But if x ∈ Q′ and y ∈ Q we must have that |x− y| ≤ 2√

nℓ(Q), which implies that

|x− y|−n ≥ (2√

n)−n|Q|−1. Let Cn = Γ ( n+12)(2√

n)−nπ−n+1

2 . It follows that for all

0 < α <Cn AvgQ f we have

Q′ ⊆

x ∈ Rn :∣∣

n

∑j=1

R j( f )(x)∣∣> α

.

Since the operator ∑nj=1 R j is of weak type (p, p) with respect to w (with constant

C), we must have

w(Q′)≤ Cp

α p

∫

Qf (x)pw(x)dx

for all α <Cn AvgQ f , which implies that

(Avg

Q

f)p ≤ C

−pn Cp

w(Q′)

∫

Qf (x)pw(x)dx . (7.4.20)

We observe that we can reverse the roles of Q and Q′ and obtain

(Avg

Q′g)p ≤ C

−pn Cp

w(Q)

∫

Q′g(x)pw(x)dx (7.4.21)

for all g supported in Q′. In particular, taking g = χQ′ in (7.4.21) gives that

w(Q)≤C−pn Cpw(Q′) .

Using this estimate and (7.4.20), we obtain

(Avg

Q

f)p ≤ (C−p

n Cp)2

w(Q)

∫

Qf (x)pw(x)dx . (7.4.22)

Using the characterization of the Ap characteristic constant in Proposition 7.1.5 (8),

it follows that

[w]Ap ≤ (C−pn Cp)2 < ∞ ;

hence w ∈ Ap.

Exercises

7.4.1. Let 1 ≤ p < ∞ and let w ∈ L1loc(R

n) satisfy w > 0 a.e. Show that C ∞0 (Rn) is

dense in Lp(w). In particular this assertion holds for any w ∈ A∞.

7.4.2. ([74]) Let T be in CZO(δ ,A,B). Show that for all ε > 0 and all 1 < p < ∞there exists a constant Cn,p,ε ,δ such that for all f ∈ Lp(Rn) and for all measurable

nonnegative functions u with u1+ε ∈ L1loc(R

n) and M(u1+ε)< ∞ a.e. we have

∫

Rn|T (∗)( f )|p udx≤Cn,p,ε ,δ (A+B)p

∫

Rn| f |pM(u1+ε)

11+ε dx .

[Hint: Obtain this result as a consequence of Theorems 7.4.6 and 7.2.7.

]

7.4.3. Use the idea of the proof of Theorem 7.4.6 to prove the following result.

Suppose that for some fixed A,B > 0 the nonnegative µ-measurable functions F

and G on a σ -finite measure space (X ,µ) satisfy the distributional inequality

µ(G > α∩F ≤ cα

)≤ Aµ

(G > Bα

)

for all α > 0. Given 0 0, w ∈ A1, and f ∈ L1(Rn,w) ∩ L1(Rn). Let f = g + b be the

Calderon–Zygmund decomposition of f at height α > 0 given in Theorem 5.3.1,

such that b=∑ j b j, where each b j is supported in a dyadic cube Q j,∫

Q jb j(x)dx= 0,

and Q j and Qk have disjoint interiors when j = k. Prove that

(a) ‖g‖L1(w) ≤ [w]A1‖ f‖L1(w) and ‖g‖L∞(w) = ‖g‖L∞ ≤ 2nα ,

(b) ‖b j‖L1(w) ≤ (1+[w]A1)‖ f‖L1(Q j ,w)

and ‖b‖L1(w) ≤ (1+[w]A1)‖ f‖L1(w),

(c) ∑ j w(Q j)≤[w]A1α ‖ f‖L1(w).

7.4.5. Assume that T is an operator associated with a kernel in SK(δ ,A). Suppose

that T maps L2(w) to L2(w) for all w ∈ A1 with bound Bw. Prove that there is a

constant Cn,δ such that

‖T‖L1(w)→L1,∞(w) ≤Cn,δ (A+Bw) [w]2A1

for all w ∈ A1.[Hint: Apply the idea of the proof of Theorem 5.3.3 using the Calderon-Zygmund

decomposition f = g+b of Exercise 7.4.4 at height γα for a suitable γ . To estimate

T (g) use an L2(w) estimate and Exercise 7.4.4. To estimate T (b) use the mean value

property, the fact that

∫

Rn\Q∗j

|y− c j|δ|x− c j|n+δ

w(x)dx≤Cδ ,nM(w)(y)≤C′δ ,n[w]A1w(y) ,

and Exercise 7.4.4 to obtain the required estimate.]

7.4.6. Recall that the transpose T t of a linear operator T is defined by

⟨T ( f ),g

⟩=

⟨f ,T t(g)

⟩

for all suitable f and g. Suppose that T is a linear operator that maps Lp(Rn,vdx)to itself for some 1 < p < ∞ and some v ∈ Ap. Show that the transpose operator T t

maps Lp′(Rn,wdx) to itself with the same norm, where w = v1−p′ ∈ Ap′ .

7.4.7. Suppose that T is a linear operator that maps L2(Rn,vdx) to itself for all v

such that v−1 ∈ A1. Show that the transpose operator T t of T maps L2(Rn,wdx) to

itself for all w ∈ A1.

7.4.8. Let 1 < p < ∞. Suppose that T is a linear operator that maps Lp(v) to itself

for all v satisfying v−1 ∈ Ap. Show that the transpose operator T t of T maps Lp′(w)to itself for all w satisfying w−1 ∈ Ap′ .

7.5 Further Properties of Ap Weights

In this section we discuss other properties of Ap weights. Many of these proper-

ties indicate deep connections with other branches of analysis. We focus attention

on three such properties: factorization, extrapolation, and relations of weighted in-

equalities to vector-valued inequalities.

7.5.1 Factorization of Weights

Recall the simple fact that if w1,w2 are A1 weights, then w = w1 w1−p2 is an Ap

weight (Exercise 7.1.2). The factorization theorem for weights says that the converse

of this statement is true. This provides a surprising and striking representation of Ap

weights.

Theorem 7.5.1. Suppose that w is an Ap weight for some 1 < p < ∞. Then there

exist A1 weights w1 and w2 such that

w = w1 w1−p2 .

7.5 Further Properties of Ap Weights 547

Proof. Let us fix a p≥ 2 and w ∈ Ap. We define an operator T as follows:

T ( f ) =(w− 1

p M( f p−1w1p )) 1

p−1 +w1p M( f w

− 1p ) ,

where M is the Hardy–Littlewood maximal operator. We observe that T is well

defined and bounded on Lp(Rn). This is a consequence of the facts that w− 1

p−1

is an Ap′ weight and that M maps Lp′(w−1

p−1 ) to itself and also Lp(w) to itself.

Thus the norm of T on Lp depends only on the Ap characteristic constant of w.

Let B(w) = ‖T‖Lp→Lp , the norm of T on Lp. Next, we observe that for f ,g ≥ 0 in

Lp(Rn) and λ ≥ 0 we have

T ( f +g)≤ T ( f )+T (g) , T (λ f ) = λT ( f ) . (7.5.1)

To see the first assertion, we need only note that for every ball B, the operator

f →(

1

|B|

∫

B| f |p−1w

1p dx

) 1p−1

is sublinear as a consequence of Minkowski’s integral inequality, since p−1≥ 1.

We now fix an Lp function f0 with ‖ f0‖Lp = 1 and we define a function ϕ in

Lp(Rn) as the sum of the Lp convergent series

ϕ =∞

∑j=1

(2B(w))− jT j( f0) . (7.5.2)

We define

w1 = w1pϕ p−1 , w2 = w

− 1pϕ ,

so that w = w1 w1−p2 . It remains to show that w1,w2 are A1 weights. Applying T and

using (7.5.1), we obtain

T (ϕ) ≤ 2B(w)∞

∑j=1

(2B(w))− j−1T j+1( f0)

= 2B(w)(ϕ− T ( f0)

2B(w)

)

≤ 2B(w)ϕ ,

that is, (w− 1

p M(ϕ p−1w1p )) 1

p−1 +w1p M(ϕw

− 1p )≤ 2B(w)ϕ .

Using that ϕ = (w−1p w1)

1p−1 = w

1p w2, we obtain

M(w1)≤ (2B(w))p−1w1 and M(w2)≤ 2B(w)w2 .

These show that w1 and w2 are A1 weights whose characteristic constants depend on

[w]Ap (and also the dimension n and p). This concludes the case p≥ 2.

We now turn to the case p < 2. Given a weight w∈ Ap for 1 2, using the result we obtained, we

write w−1/(p−1) = v1v1−p′2 , where v1, v2 are A1 weights. It follows that w = v

1−p1 v2,

and this completes the asserted factorization of Ap weights.

Combining the result just obtained with Theorem 7.2.7, we obtain the following

description of Ap weights.

Corollary 7.5.2. Let w be an Ap weight for some 1 < p <∞. Then there exist locally

integrable functions f1 and f2 with

M( f1)+M( f2)< ∞ a.e.,

constants 0 < ε1,ε2 < 1, and a nonnegative function k satisfying k,k−1 ∈ L∞ such

that

w = k M( f1)ε1M( f2)

ε2(1−p) . (7.5.3)

7.5.2 Extrapolation from Weighted Estimates on a Single Lp0

Our next topic concerns a striking application of the class of Ap weights. It says

that an estimate on Lp0(v) for a single p0 and all Ap0weights v implies a similar Lp

estimate for all p in (1,∞). This property is referred to as extrapolation.

Surprisingly the operator T is not needed to be linear or sublinear in the following

extrapolation theorem. The only condition required is that T be well defined on⋃1≤q<∞

⋃w∈Aq

Lq(w). If T happens to be a linear operator, this condition can be

relaxed to T being well defined on C ∞0 (Rn).

Theorem 7.5.3. Suppose that T is defined on⋃

1≤q<∞

⋃w∈Aq

Lq(w) and takes values

in the space of measurable complex-valued functions. Let 1 ≤ p0 < ∞ and suppose

that there exists a positive increasing function N on [1,∞) such that for all weights

v in Ap0we have ∥∥T

∥∥Lp0 (v)→Lp0 (v)

≤ N([v]Ap0

). (7.5.4)

Then for any 1 < p < ∞ and for all weights w in Ap we have

∥∥T∥∥

Lp(w)→Lp(w)≤ K

(n, p, p0, [w]Ap

), (7.5.5)

where

K(n, p, p0, [w]Ap

)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩

2N(κ1(n, p, p0) [w]

p0−1

p−1

Ap

)when p < p0,

2p−p0

p0(p−1) N(κ2(n, p, p0) [w]Ap

)when p > p0,

and κ1(n, p, p0) and κ2(n, p, p0) are constants that depend on n, p, and p0.

Proof. Let 1 < p < ∞ and w ∈ Ap. We define an operator

M′( f ) =M( f w)

w,

where M is the Hardy–Littlewood maximal operator. We observe that since w1−p′ is

in Ap′ , the operator M′ maps Lp′(w) to itself; indeed, we have

∥∥M′∥∥Lp′ (w)→Lp′ (w) =

∥∥M∥∥

Lp′ (w1−p′ )→Lp′ (w1−p′ )

≤ Cn,p[w1−p′ ]

1p′−1

Ap′

= Cn,p[w]Ap

(7.5.6)

in view of Theorem 7.1.9 and property (4) of Proposition 7.1.5.

We introduce operators M0( f ) = | f | and Mk = M M · · · M, where M is the

Hardy–Littlewood maximal function and the composition is taken k times. Likewise,

we introduce powers (M′)k of M′ for k ∈ Z+ ∪0. The following lemma provides

the main tool in the proof of Theorem 7.5.3. Its simple proof uses Theorem 7.1.9 and

(7.5.6) and is omitted.

Lemma 7.5.4. Let 1 < p < ∞ and w ∈ Ap. Define operators R and R′

R( f ) =∞

∑k=0

Mk( f )(2‖M‖Lp(w)→Lp(w)

)k

for functions f in Lp(w) and also

R′( f ) =∞

∑k=0

(M′)k( f )(2∥∥M′

∥∥Lp′ (w)→Lp′ (w)

)k

for functions f in Lp′(w). Then there exist constants C1(n, p) and C2(n, p) that de-

pend on n and p such that

| f | ≤ R( f ) , (7.5.7)∥∥R( f )∥∥

Lp(w)≤ 2

∥∥ f∥∥

Lp(w), (7.5.8)

M(R( f )) ≤ C1(n, p) [w]1

p−1

ApR( f ) , (7.5.9)

for all functions f in Lp(w) and such that

|h| ≤ R′(h) , (7.5.10)∥∥R′(h)∥∥

Lp′ (w) ≤ 2∥∥h

∥∥Lp′ (w) , (7.5.11)

M′(R′(h)) ≤ C2(n, p) [w]Ap R′(h) , (7.5.12)

for all functions h in Lp′(w).

We now proceed with the proof of the theorem. It is natural to split the proof into

the cases p < p0 and p > p0.

Case (1): p < p0. Assume momentarily that R( f )− p0

(p0/p)′ is an Ap0weight. Then we

have

∥∥T ( f )∥∥p

Lp(w)

=∫

Rn|T ( f )|pR( f )

− p

(p0/p)′ R( f )p

(p0/p)′ wdx

≤(∫

Rn|T ( f )|p0R( f )

− p0(p0/p)′ wdx

) pp0

(∫

RnR( f )p wdx

) 1(p0/p)′

≤ N([

R( f )− p0

(p0/p)′]

Ap0

)p(∫

Rn| f |p0R( f )

− p0(p0/p)′ wdx

) pp0

(∫

RnR( f )p wdx

) 1(p0/p)′

≤ N([

R( f )− p0

(p0/p)′]

Ap0

)p(∫

RnR( f )p0R( f )

− p0(p0/p)′ wdx

) pp0

(∫

RnR( f )p wdx

) 1(p0/p)′

= N([

R( f )− p0

(p0/p)′]

Ap0

)p(∫

RnR( f )pwdx

) pp0

(∫

RnR( f )p wdx

) 1(p0/p)′

≤ N([

R( f )− p0

(p0/p)′]

Ap0

)p(2∥∥ f

∥∥Lp(w)

)p,

where we used Holder’s inequality with exponents p0/p and (p0/p)′, the hypothesis

of the theorem, (7.5.7), and (7.5.8). Thus, we have the estimate

∥∥T ( f )∥∥

Lp(w)≤ 2N

([R( f )

− p0(p0/p)′

]Ap0

)∥∥ f∥∥

Lp(w)(7.5.13)

and it remains to obtain a bound for the Ap0characteristic constant of R( f )

− p0(p0/p)′ .

In view of (7.5.9), the function R( f ) is an A1 weight with characteristic constant at

most a constant multiple of [w]1

p−1

Ap. Consequently, there is a constant C′1 such that

R( f )−1 ≤C′1 [w]1

p−1

Ap

(1

|Q|

∫

QR( f )dx

)−1

for any cube Q in Rn. Thus we have

1

|Q|

∫

QR( f )

− p0(p0/p)′ wdx

≤(C′1 [w]

1p−1

Ap

) p0(p0/p)′

(1

|Q|

∫

QR( f )dx

)− p0(p0/p)′

(1

|Q|

∫

Qwdx

).

(7.5.14)

Next we have

(1

|Q|

∫

Q

(R( f )

− p0(p0/p)′ w

)1−p′0dx

)p0−1

=

(1

|Q|

∫

QR( f )

p0(p′0−1)

(p0/p)′ w1−p′0 dx

)p0−1

≤(

1

|Q|

∫

QR( f )dx

) p0(p0/p)′

(1

|Q|

∫

Qw1−p′

)p−1

,

(7.5.15)

where we applied Holder’s inequality with exponents

(p′−1

p′0−1

)′and

p′−1

p′0−1,

and we used that

p0(p′0−1)

(p0/p)′

(p′−1

p′0−1

)′= 1 and

p0−1(p′−1

p′0−1

)′ =p0

(p0/p)′.

Multiplying (7.5.14) by (7.5.15) and taking the supremum over all cubes Q in Rn we

deduce that

[R( f )

− p0(p0/p)′

]Ap0

≤(C′1 [w]

1p−1

Ap

) p0(p0/p)′ [w]Ap = κ1(n, p, p0) [w]

p0−1

p−1

Ap.

Combining this estimate with (7.5.13) and using the fact that N is an increasing

function, we obtain the validity of (7.5.5) in the case p < p0.

Case (2): p > p0. In this case we set r = p/p0 > 1. Then we have

∥∥T ( f )∥∥p

Lp(w)=

∥∥ |T ( f )|p0∥∥r

Lr(w)=

(∫

Rn|T ( f )|p0hwdx

)r

(7.5.16)

for some nonnegative function h with Lr′(w) norm equal to 1. We define a function

H =[R′(h

r′p′)] p′

r′ .

Obviously, we have 0≤ h≤ H and thus

∫

Rn|T ( f )|p0hwdx ≤

∫

Rn|T ( f )|p0H wdx

≤ N([H w]Ap0

)p0∥∥ f

∥∥p0

Lp0 (H w)

≤ N([H w]Ap0

)p0∥∥ | f |p0

∥∥Lr(w)

∥∥H∥∥

Lr′ (w)

≤ 2p′r′ N

([H w]Ap0

)p0∥∥ f

∥∥p0

Lp(w),

(7.5.17)


noting that

∥∥H∥∥r′

Lr′ (w) =∫

RnR′(hr′/p′)p′ wdx≤ 2p′

∫

Rnhr′ wdx = 2p′ ,

which is valid in view of (7.5.11). Moreover, this argument is based on the hypothesis

of the theorem and requires that H w be an Ap0weight. To see this, we observe that

condition (7.5.12) implies that Hr′/p′w is an A1 weight with characteristic constant

at most a multiple of [w]A1. Thus, there is a constant C′2 that depends only on n and

p such that1

|Q|

∫

QH

r′p′ wdx≤C′2 [w]ApH

r′p′ w

for all cubes Q in Rn. From this it follows that

(H w)−1 ≤ κ2(n, p, p0) [w]p′r′Ap

(1

|Q|

∫

QH

r′p′ wdx

)− p′r′

wp′r′ −1 ,

where we set κ2(n, p, p0) = (C′2)p′/r′ . We raise the preceding displayed expression

to the power p′0− 1, we average over the cube Q, and then we raise to the power

p0−1. We deduce the estimate

(1

|Q|

∫

Q(H w)1−p′0 dx

)p0−1

≤ κ2(n, p, p0) [w]p′r′Ap

(1

|Q|

∫

QH

r′p′ wdx

)− p′r′(

1

|Q|

∫

Qw1−p′ dx

)p0−1

,

(7.5.18)

where we use the fact that

(p′

r′−1

)(p′0−1

)= 1− p′ .

Note that r′/p′ ≥ 1, since p0 ≥ 1. Using Holder’s inequality with exponents r′/p′

and (r′/p′)−1 we obtain that

1

|Q|

∫

QH wdx≤

(1

|Q|

∫

QH

r′p′ wdx

) p′r′(

1

|Q|

∫

Qwdx

) p0−1

p−1

, (7.5.19)

where we used that1

( r′p′ )′ =

p0−1

p−1.

Multiplying (7.5.18) by (7.5.19), we deduce the estimate

[H w

]Ap0

≤ κ2(n, p, p0) [w]p′r′Ap

[w]p0−1

p−1

Ap= κ2(n, p, p0) [w]Ap .

Inserting this estimate in (7.5.17) we obtain

∫

Rn|T ( f )|p0hwdx≤ 2

p′r′ N

(κ2(n, p, p0) [w]Ap

)p0∥∥ f

∥∥p0

Lp(w),

and combining this with (7.5.16) we conclude that

∥∥T ( f )∥∥p

Lp(w)≤ 2

p′rr′ N

(κ2(n, p, p0) [w]Ap

)p0r∥∥ f∥∥p0r

Lp(w).

This proves the required estimate (7.5.5) in the case p > p0.

There is a version of Theorem 7.5.3 in which the initial strong type assumption

is replaced by a weak type estimate.

Theorem 7.5.5. Suppose that T is a well defined operator on⋃

1<q<∞

⋃w∈Aq

Lq(w)that takes values in the space of measurable complex-valued functions. Fix 1 ≤p0 <∞ and suppose that there is an increasing function N on [1,∞) such that for all

weights v in Ap0we have

∥∥T∥∥

Lp0 (v)→Lp0,∞(v)≤ N([v]Ap0

) . (7.5.20)

Then for any 1 0 we define

Tλ ( f ) = λχ|T ( f )|>λ .

The operator Tλ is not linear but is well defined on⋃

1<q<∞

⋃w∈Aq

Lq(w), since T is

well defined on this union. We show that Tλ maps Lp0(v) to Lp0(v) for every v∈ Ap0.

Indeed, we have

(∫

Rn|Tλ ( f )|p0 vdx

) 1p0

=

(∫

Rnλ p0χ|T ( f )|>λ vdx

) 1p0

=(λ p0 v

(|T ( f )|> λ

)) 1p0

≤ N([v]Ap0)∥∥ f

∥∥Lp0 (v)

using the hypothesis on T . Applying Theorem 7.5.3, we obtain that Tλ maps Lp(w)to itself for all 1 < p <∞ and all w∈ Ap with a constant independent of λ . Precisely,

for any w ∈ Ap and any f ∈ Lp(w) we have

∥∥Tλ ( f )∥∥

Lp(w)≤ K

(n, p, p0, [w]Ap

)∥∥ f∥∥

Lp(w).

Since ∥∥T ( f )∥∥

Lp,∞(w)= sup

λ>0

∥∥Tλ ( f )∥∥

Lp(w),

it follows that T maps Lp(w) to Lp,∞(w) with the asserted norm.

Assuming that the operator T in the preceding theorem is sublinear (or quasi-

sublinear), we obtain the following result that contains a stronger conclusion.

Corollary 7.5.6. Suppose that T is a sublinear operator on⋃

1<q<∞

⋃w∈Aq

Lq(w)that takes values in the space of measurable complex-valued functions. Fix 1 ≤p0 <∞ and suppose that there is an increasing function N on [1,∞) such that for all

weights v in Ap0we have

∥∥T∥∥

Lp0 (v)→Lp0,∞(v)≤ N([v]Ap0

) . (7.5.22)

Then for any 1< p<∞ and any weight w in Ap there is a constant K′(n, p, p0, [w]Ap)such that ∥∥T ( f )

∥∥Lp(w)

≤ K′(n, p, p0, [w]Ap)∥∥ f

∥∥Lp(w)

.

Proof. The proof follows from Theorem 7.5.5 and the Marcinkiewicz interpolation

theorem.

We end this subsection by observing that the conclusion of the extrapolation The-

orem 7.5.3 can be strengthened to yield vector-valued estimates. This strengthening

may be achieved by a simple adaptation of the proof discussed.

Corollary 7.5.7. Suppose that T is defined on⋃

1≤q<∞

⋃w∈Aq

Lq(w) and takes val-

ues in the space of all measurable complex-valued functions. Fix 1 ≤ p0 < ∞ and

suppose that there is an increasing function N on [1,∞) such that for all weights v

in Ap0we have ∥∥T

∥∥Lp0 (v)→Lp0 (v)

≤ N([v]Ap0

).

Then for every 1 < p < ∞ and every weight w ∈ Ap we have

∥∥∥(∑

j

|T ( f j)|p0

) 1p0

∥∥∥Lp(w)

≤ K(n, p, p0, [w]Ap)∥∥∥(∑

j

| f j|p0

) 1p0

∥∥∥Lp(w)

for all sequences of functions f j in Lp(w), where K(n, p, p0, [w]Ap

)is as in Theorem

7.5.3.

Proof. To derive the claimed vector-valued inequality follow the proof of Theorem

7.5.3 replacing the function f by (∑ j | f j|p0)1

p0 and T ( f ) by (∑ j |T ( f j)|p0)1

p0 .

7.5.3 Weighted Inequalities Versus Vector-Valued Inequalities

We now discuss connections between weighted inequalities and vector-valued

inequalities. The next result provides strong evidence that there is a nontrivial

connection of this sort. The following is a general theorem saying that any vector-

valued inequality is equivalent to some weighted inequality. The proof of the the-

orem is based on a minimax lemma whose precise formulation and proof can be

found in Appendix H.

Theorem 7.5.8. (a) Let 0 < p < q,r < ∞. Let Tj j be a sequence of sublinear

operators that map Lq(µ) to Lr(ν), where µ and ν are arbitrary measures. Then

the vector-valued inequality

∥∥∥(∑

j

|Tj( f j)|p)1

p

∥∥∥Lr≤C

∥∥∥(∑

j

| f j|p)1

p

∥∥∥Lq

(7.5.23)

holds for all f j ∈ Lq(µ) if and only if for every u≥ 0 in Lr

r−p (ν) there exists U ≥ 0

in Lq

q−p (µ) with

‖U‖L

qq−p

≤ ‖u‖L

rr−p

,

supj

∫|Tj( f )|p udν ≤ Cp

∫| f |p U dµ .

(7.5.24)

(b) Let 0 < q,r < p < ∞. Let Tj j be as before. Then the vector-valued inequality

(7.5.23) holds for all f j ∈ Lq(µ) if and only if for every u≥ 0 in Lq

p−q (µ) there exists

U ≥ 0 in Lr

p−r (ν) with

‖U‖L

rp−r

≤ ‖u‖L

qp−q

,

supj

∫|Tj( f )|p U−1 dν ≤ Cp

∫| f |p u−1 dµ .

(7.5.25)

Proof. We begin with part (a). Given f j ∈ Lq(Rn,µ), we use (7.5.24) to obtain

∥∥∥(∑

j

|Tj( f j)|p) 1

p∥∥∥

Lr(ν)=

∥∥∥∑j

|Tj( f j)|p∥∥∥

1p

Lrp (ν)

= sup‖u‖

Lr

r−p≤1

(∫

Rn∑

j

|Tj( f j)|p udν

)1p

≤ sup‖u‖

Lr

r−p≤1

C

(∫

Rn∑

j

| f j|p U dµ

)1p

≤ sup‖u‖

Lr

r−p≤1

C

∥∥∥∑j

| f j|p∥∥∥

1p

Lqp (µ)

‖U‖1p

Lq

q−p

≤ C

∥∥∥(∑

j

| f j|p) 1

p∥∥∥

Lq(µ),


which proves (7.5.23) with the same constant C as in (7.5.24). To prove the converse,

given a nonnegative u ∈ Lr

r−p (ν) with ‖u‖L

rr−p

= 1, we define

A =

a = (a0,a1) : a0 =∑j

| f j|p, a1 =∑j

|Tj( f j)|p, f j ∈ Lq(µ)

and

B =

b ∈ Lq

q−p (µ) : b≥ 0 , ‖b‖L

qq−p≤ 1 = ‖u‖

Lr

r−p

.

Notice that A and B are convex sets and B is weakly compact. (The sublinearity of

each Tj is used here.) We define the function Φ on A×B by setting

Φ(a,b) =∫

a1udν−Cp

∫a0bdµ =∑

j

(∫|Tj( f j)|pudν−Cp

∫| f j|pbdµ

).

Then Φ is concave on A and weakly continuous and convex on B. Thus the minimax

lemma in Appendix H is applicable. This gives

minb∈B

supa∈A

Φ(a,b) = supa∈A

minb∈B

Φ(a,b) . (7.5.26)

At this point observe that for a fixed a =(∑ j | f j|p,∑ j |Tj( f j)|p

)in A we have

minb∈B

Φ(a,b) ≤∥∥∥∑

j

|Tj( f j)|p∥∥∥

Lrp (ν)‖u‖

Lr

r−p−Cp max

b∈B

∫∑

j

| f j|p bdµ

≤∥∥∥∑

j

|Tj( f j)|p∥∥∥

Lrp (ν)

−Cp∥∥∥∑

j

| f j|p∥∥∥

Lqp (µ)

≤ 0

using the hypothesis (7.5.23). It follows that supa∈A minb∈BΦ(a,b) ≤ 0 and hence

(7.5.26) yields minb∈B supa∈AΦ(a,b) ≤ 0. Thus there exists a U ∈ B such that

Φ(a,U)≤ 0 for every a ∈ A. This completes the proof of part (a).

The proof of part (b) is similar. Using the result of Exercise 7.5.1 and (7.5.25),

given f j ∈ Lq(Rn,µ) we have

∥∥∥(∑

j

| f j|p) 1

p

∥∥∥Lq(µ)

=∥∥∥∑

j

| f j|p∥∥∥

1p

Lqp (µ)

= inf‖u‖

L

qp−q

≤1

(∫

Rn∑

j

| f j|p u−1 dµ

)1p

≥ 1

Cinf

‖U‖L

rp−r

≤1

(∫

Rn∑

j

|Tj( f j)|p U−1 dν

)1p

=1

C

∥∥∥∑j

|Tj( f j)|p∥∥∥

1p

Lrp (ν)

=1

C

∥∥∥(∑

j

|Tj( f j)|p) 1

p∥∥∥

Lr(ν).

To prove the converse direction in part (b), given a fixed u ≥ 0 in Lq

p−q (µ) with∥∥u∥∥

Lq

p−q= 1, we define A as in part (a) and

B =

b ∈ Lp

p−r (ν) : b≥ 0, ‖b‖L

pp−r≤ 1 = ‖u‖

Lq

p−q

.

We also define the function Φ on A×B by setting

Φ(a,b) =∫

a1b−1dν−Cp

∫a0u−1 dµ

= ∑j

(∫|Tj( f j)|pb−1 dν−Cp

∫| f j|pu−1 dµ

).

Then Φ is concave on A and weakly continuous and convex on B. Also, using Exer-

cise 7.5.1, for any a =(∑ j | f j|p,∑ j |Tj( f j)|p

)in A, we have

minb∈B

Φ(a,b)≤∥∥∥∑

j

|Tj( f j)|p∥∥∥

Lrp (ν)

−Cp∥∥∥∑

j

| f j|p∥∥∥

Lqp (µ)

≤ 0 .

Thus supa∈A minb∈BΦ(a,b) ≤ 0. Using (7.5.26), yields minb∈B supa∈AΦ(a,b) ≤ 0,

and the latter implies the existence of a U in B such that Φ(a,U)≤ 0 for all a ∈ A.

This proves (7.5.25).

Example 7.5.9. We use the previous theorem to obtain another proof of the vector-

valued Hardy–Littlewood maximal inequality in Corollary 5.6.5. We take Tj =M for

all j. For given 1< p< q<∞ and u in Lq

q−p we set s= qq−p

and U = ‖M‖−1Ls→Ls M(u).

In view of Exercise 7.1.7 we have

‖U‖Ls ≤ ‖u‖Ls and

∫

RnM( f )p udx≤Cp

∫

Rn| f |p U dx .

Using Theorem 7.5.8, we obtain

∥∥∥(∑

j

|M( f j)|p)1

p

∥∥∥Lq≤Cn,p,q

∥∥∥(∑

j

| f j|p)1

p

∥∥∥Lq

(7.5.27)

whenever 1 < p < q < ∞, an inequality obtained earlier in (5.6.25).

It turns out that no specific properties of the Hardy–Littlewood maximal function

were used in the preceding inequality, and one could obtain a general result along

these lines.

Exercises

7.5.1. Let (X ,µ) be a measure space, 0 < s < 1, and f ∈ Ls(X ,µ). Show that

∥∥ f∥∥

Ls = inf

∫

X| f |u−1 dµ : ‖u‖

Ls

1−s≤ 1

and that the infimum is attained.[Hint: Try u = c | f |1−s for a suitable constant c.

]

7.5.2. (K. Yabuta) Let w ∈ Ap for some 1 < p < ∞ and let f be in Lploc(R

n,wdx).Show that f lies in L1

loc(Rn).[

Hint: Write w = w1/wp−12 via Theorem 7.5.1.

]

7.5.3. Use the same idea of the proof of Theorem 7.5.1 to prove the following gen-

eral result: Let µ be a positive measure on a measure space X and let T be a bounded

sublinear operator on Lp(X ,µ) for some 1≤ p <∞. Suppose that T ( f )≥ 0 for all f

in Lp(X ,µ). Prove that for all f0 ∈ Lp(X ,µ), there exists an f ∈ Lp(X ,µ) such that

(a) f0(x)≤ f (x) for µ-almost all x ∈ X .

(b) ‖ f‖Lp(X) ≤ 2‖ f0‖Lp(X).

(c) T ( f )(x)≤ 2‖T‖Lp→Lp f (x) for µ-almost all x ∈ X .[Hint: Try the expression in (7.5.2) starting the sum at j = 0.

]

7.5.4. ([100]) Suppose that T is an operator defined on⋃

1<q<∞

⋃w∈Aq

Lq(w) that

satisfies ‖T‖Lr(v)→Lr(v) ≤ N([v]Ar) for some increasing function N : [1,∞)→ R+.

Without using Theorem 7.5.3 prove that for 1 < q < r and all v ∈ A1, T maps Lq(v)to Lq(v) with constant depending on q,r, n, and [v]A1

.[Hint: Holder’s inequality gives that

∥∥T ( f )∥∥

Lq(v)≤

(∫

Rn|T ( f )(x)|rM( f )(x)q−r v(x)dx

)1r(∫

RnM( f )(x)qv(x)dx

)r−qrq

.

Then use the fact that the weight M( f )r−qr−1 is in A1 and Exercise 7.1.2.

]

7.5.5. Let T be a sublinear operator defined on⋃

2≤q<∞Lq. Suppose that for all

functions f and u we have

∫

Rn|T ( f )|2udx≤

∫

Rn| f |2M(u)dx .

Prove that T maps Lp(Rn) to itself for all 2 < p < ∞.[Hint: Use that

∥∥T ( f )∥∥

Lp = sup‖u‖

L(p/2)′ ≤1

(∫

Rn|T ( f )|2udx

)12

and Holder’s inequality.]

7.5.6. (X. C. Li) Let T be a sublinear operator defined on⋃

1<q≤2

⋃w∈Aq

Lq(w).

Suppose that T maps L2(w) to L2(w) for all weights w that satisfy w−1 ∈ A1. Prove

that T maps Lp to itself for all 1 < p < 2.[Hint: We have

∥∥T ( f )∥∥

Lp ≤(∫

Rn|T ( f )|2M( f )−(2−p) dx

)12(∫

RnM( f )p dx

)2−p2p

by Holder’s inequality. Apply the hypothesis to the first term of the product.]

HISTORICAL NOTES

Weighted inequalities can probably be traced back to the beginning of integration, but the

Ap condition first appeared in a paper of Rosenblum [298] in a somewhat different form. The

characterization of Ap when n = 1 in terms of the boundedness of the Hardy–Littlewood maximal

operator was obtained by Muckenhoupt [260]. The estimate on the norm in (7.1.25) can also be

reversed, as shown by Buckley [38]. The simple proof of Theorem 7.1.9 is contained in Lerner’s

article [218] and yields both the Muckenhoupt theorem and Buckley’s optimal growth of the norm

of the Hardy–Littlewood maximal operator in terms of the Ap characteristic constant of the weight.

Another proof of this result is given by Christ and Fefferman [61]. Versions of Lemma 7.1.10 for

balls were first obtained by Besicovitch [27] and independently by Morse [258]. The particular

version of Lemma 7.1.10 that appears in the text is adapted from that in de Guzman [93]. Another

version of this lemma is contained in the book of Mattila [246]. The fact that A∞ is the union

of the Ap spaces was independently obtained by Muckenhoupt [261] and Coifman and Fefferman

[66]. The latter paper also contains a proof that Ap weights satisfy the crucial reverse Holder

condition. This condition first appeared in the work of Gehring [125] in the following context: If F

is a quasiconformal homeomorphism from Rn into itself, then |det(∇F)| satisfies a reverse Holder

inequality. The characterization of A1 weights is due to Coifman and Rochberg [68]. The fact that

M( f )δ is in A∞ when δ < 1 was previously obtained by Cordoba and Fefferman [74]. The different

characterizations of A∞ (Theorem 7.3.3) are implicit in [260] and [66]. Another characterization of

A∞ in terms of the Gurov-Reshetnyak condition supQ1|Q|

∫Q | f −AvgQ f |dx ≤ εAvgQ f for f ≥ 0

and 0 < ε < 2 was obtained by Korenovskyy, Lerner, and Stokolos [201]. The definition of A∞using the reverse Jensen inequality herein was obtained as an equivalent characterization of that

space by Garcıa-Cuerva and Rubio de Francia [122] (p. 405) and independently by Hruscev [161].

The reverse Holder condition was extensively studied by Cruz-Uribe and Neugebauer [82].

Weighted inequalities with weights of the form |x|a for the Hilbert transform were first obtained

by Hardy and Littlewood [147] and later by Stein [332] for other singular integrals. The necessity

and sufficiency of the Ap condition for the boundedness of the Hilbert transform on weighted Lp

spaces was obtained by Hunt, Muckenhoupt, and Wheeden [167]. Historically, the first result re-

lating Ap weights and the Hilbert transform is the Helson-Szego theorem [149], which says that

the Hilbert transform is bounded on L2(w) if and only if logw = u+Hv, where u,v ∈ L∞(R) and

‖v‖L∞ < π2

. The Helson-Szego condition easily implies the A2 condition, but the only known direct

proof for the converse gives ‖v‖L∞ < π; see Coifman, Jones, and Rubio de Francia [67]. A related

result in higher dimensions was obtained by Garnett and Jones [123]. Weighted Lp estimates con-

trolling Calderon–Zygmund operators by the Hardy–Littlewood maximal operator were obtained

by Coifman [65]. Coifman and Fefferman [66] extended one-dimensional weighted norm inequal-

ities to higher dimensions and also obtained good lambda inequalities for A∞ weights for more

general singular integrals and maximal singular integrals (Theorem 7.4.3). Bagby and Kurtz [19],

and later Alvarez and Perez [4], gave a sharper version of Theorem 7.4.3, by replacing the good

lambda inequality by a rearrangement inequality. See also the related work of Lerner [217]. The

following relation ‖Md( f )‖Lp(w) ≤C(p,n, [w]A∞)‖M#( f )‖Lp(w) between the dyadic maximal func-

tion and the sharp maximal function is valid for any w ∈ A∞ under the condition M( f ) ∈ Lp0 but

also under the weaker assumption that w(| f | > t) < ∞ for every t > 0; see Kurtz [208]. Using

that min(M,w) is an A∞ weight with constant independent of M and Fatou’s lemma, this condition

can be relaxed to || f |> t|<∞ for every t > 0. A rearrangement inequality relating f and M#( f )is given in Bagby and Kurtz [18].

The factorization of Ap weights was conjectured by Muckenhoupt and proved by Jones [179].

The simple proof given in the text can be found in [67]. Extrapolation of operators (Theorem 7.5.3)

is due to Rubio de Francia [300]. An alternative proof of this theorem was given later by Garcıa-

Cuerva [121]. The value of the constant K(n, p, p0, [w]Ap ) first appeared in Dragicevic, Grafakos,

Pereyra, and Petermichl [98]. Another proof with sharp bounds (in terms of the characteristic con-

stant of the weights) was given by Duoandikoetxea [101]. The present treatment of Theorem 7.5.3,

based on crucial Lemma 7.5.4, was communicated to the author by J. M. Martell. One may also

consult the related work of Cruz-Uribe, Martell, and Perez [80]. The simple proof of Theorem 7.5.5

was conceived by J. M. Martell and first appeared in the treatment of extrapolation of operators of

many variables; see Grafakos and Martell [135]. The idea of extrapolation can be carried to general

pairs of functions, see Cruz-Uribe, Martell, and Perez [78]. Estimates for the distribution function

in extrapolation theory were obtained by Carro, Torres, and Soria [58]. The equivalence between

vector-valued inequalities and weighted norm inequalities of Theorem 7.5.8 is also due to Rubio de

Francia [299]. The difficult direction in this equivalence is obtained using a minimax principle (see

Fan [111]). Alternatively, one can use the factorization theory of Maurey [247], which brings an

interesting connection with Banach space theory. The book of Garcıa-Cuerva and Rubio de Francia

[122] provides an excellent reference on this and other topics related to weighted norm inequalities.

A primordial double-weighted norm inequality is the observation of Fefferman and Stein [115]

that the maximal function maps Lp(M(w)) to Lp(w) for nonnegative measurable functions w (Exer-

cise 7.1.7). Sawyer [312] obtained that the condition supQ

(∫Q v1−p′dx

)−1∫Q M(v1−p′χQ)

pwdx < ∞provides a characterization of all pairs of weights (v,w) for which the Hardy–Littlewood maximal

operator M maps Lp(v) to Lp(w). Simpler proofs of this result were obtained by Cruz-Uribe [77]

and Verbitsky [367]. The fact that Sawyer’s condition reduces to the usual Ap condition when v=w

was shown by Hunt, Kurtz, and Neugebauer [166]. The two-weight problem for singular integrals

is more delicate, since they are not necessarily bounded from Lp(M(w)) to Lp(w). Known results

in this direction are that singular integrals map Lp(M[p]+1(w)) to Lp(w), where Mr denotes the rth

iterate of the maximal operator. See Wilson [377] (for 1 < p < 2) and Perez [277] for the remaining

p’s. A necessary condition for the boundedness of the Hilbert transform from Lp(v) to Lp(w) was

obtained by Muckenhoupt and Wheeden [262].

For an approach to two-weighted inequalities using Bellman functions, we refer to the article of

Nazarov, Treil, and Volberg [266]. The notion of Bellman functions originated in control theory;

the article [267] of the previous authors analyzes the connections between optimal control and

harmonic analysis. Bellman functions have been used to derive estimates for the norms of classical

operators on weighted Lebesgue spaces; for instance, Petermichl [279] showed that for w ∈ A2(R),the norm of the Hilbert transform from L2(R,w) to L2(R,w) is bounded by a constant times the

characteristic constant [w]A2.

The theory of Ap weights in this chapter carries through to the situation in which Lebesgue

measure is replaced by a general doubling measure. This theory also has a substantial analogue

when the underlying measure is nondoubling but satisfies µ(∂Q) = 0 for all cubes Q in Rn with

sides parallel to the axes; see Orobitg and Perez [272]. A thorough account of weighted Littlewood–

Paley theory and exponential-square function integrability is contained in the book of Wilson [378].

The conjecture whether ‖T‖L1(M(w))→L1,∞(w) < ∞ holds for a weight w was disproved by

Reguera [287] when T is a Haar multiplier and then by Reguera and Thiele [288] for the Hilbert

transform. However, the slightly weaker version of this inequality, in which M(w) is replaced by

the Orlicz maximal operator ML(logL)ε (w), holds for any ε > 0 and any Calderon-Zygmund oper-

ator T , as shown by Perez [277]. For A1 weights w the aforementioned conjecture would imply

‖T‖L1(w)→L1,∞(w) ≤C [w]A1. However, Nazarov, Reznikov, Vasyunin, and Volberg [265] disproved

the weaker inequality ‖T‖L1(w)→L1,∞(w) ≤ C [w]A1

(log(e+ [w]A1

))α

for α < 15

. Lerner, Ombrosi,

and Perez [223] had previously shown that the preceding inequality holds with α = 1 for any

Calderon-Zygmund operator T .


Concerning the sharp weighted bound ‖T‖L2(w)→L2(w) ≤ cT [w]A2for a Calderon-Zygmund op-

erator T we have the work of Petermichl and Volberg [281] which answered a question by Astala,

Iwaniecz and Saksman [14] on the regularity of solutions to the Beltrami equation. The proofs of

this inequality for the Hilbert and Riesz transforms via the Bellman function technique were ob-

tained soon afterwards by Petermichl [279], [280]. The use of Bellman functions was first avoided

in the work of Lacey, Petermichl, and Reguera [210], whose proof recovered the already known

cases and used Haar shift operators, the two-weight theory for them of Nazarov, Treil and Volberg

[268], and corona decompositions. The simplest proof for these classical operators was obtained

by Cruz-Uribe, Martell, and Perez [79], [81] using a very powerful inequality due to Lerner [219].

The complete proof for a general Calderon-Zygmund operator was given by Hytonen [168]. A sim-

plified proof was provided by Lerner [220], [221]. For other improvements and estimates involving

Ap and A∞ constants see the work of Lerner [222], Hytonen and Perez [170], and Lacey, Hytonen,

and Perez [169].

Appendix A

Gamma and Beta Functions

A.1 A Useful Formula

The following formula is valid:

∫

Rne−|x|

2

dx =(√

π)n.

This is an immediate consequence of the corresponding one-dimensional identity

∫ +∞

−∞e−x2

dx =√π ,

which is usually proved from its two-dimensional version by switching to polar

coordinates:

I2 =∫ +∞

−∞

∫ +∞

−∞e−x2

e−y2

dydx = 2π∫ ∞

0re−r2

dr = π .

A.2 Definitions of Γ (z) and B(z,w)

For a complex number z with Rez > 0 define

Γ (z) =∫ ∞

0tz−1e−tdt.

Γ (z) is called the gamma function. It follows from its definition that Γ (z) is analytic

on the right half-plane Rez > 0.

Two fundamental properties of the gamma function are that

Γ (z+1) = zΓ (z) and Γ (n) = (n−1)! ,


DOI 10.1007/978-1-4939-1194-3, © Springer Science+Business Media New York 2014

563

564 A Gamma and Beta Functions

where z is a complex number with positive real part and n ∈ Z+. Indeed, integration

by parts yields

Γ (z) =∫ ∞

0tz−1e−t dt =

[tze−t

z

]∞

0

+1

z

∫ ∞

0tze−t dt =

1

zΓ (z+1).

Since Γ (1) = 1, the property Γ (n) = (n− 1)! for n ∈ Z+ follows by induction.

Another important fact is that

Γ(

12

)=√π .

This follows easily from the identity

Γ(

12

)=

∫ ∞

0t−

12 e−t dt = 2

∫ ∞

0e−u2

du =√π .

Next we define the beta function. Fix z and w complex numbers with positive real

parts. We define

B(z,w) =∫ 1

0tz−1(1− t)w−1 dt =

∫ 1

0tw−1(1− t)z−1 dt.

We have the following relationship between the gamma and the beta functions:

B(z,w) =Γ (z)Γ (w)

Γ (z+w),

when z and w have positive real parts.

The proof of this fact is as follows:

Γ (z+w)B(z,w) = Γ (z+w)∫ 1

0tw−1(1− t)z−1 dt

= Γ (z+w)∫ ∞

0uw−1

(1

1+u

)z+w

du t = u/(1+u)

=∫ ∞

0

∫ ∞

0uw−1

(1

1+u

)z+w

vz+w−1e−v dvdu

=∫ ∞

0

∫ ∞

0uw−1sz+w−1e−s(u+1) dsdu s = v/(1+u)

=∫ ∞

0sze−s

∫ ∞

0(us)w−1e−su duds

=∫ ∞

0sz−1e−sΓ (w)ds

= Γ (z)Γ (w) .

A.4 Computation of Integrals Using Gamma Functions 565

A.3 Volume of the Unit Ball and Surface of the Unit Sphere

We denote by vn the volume of the unit ball in Rn and by ωn−1 the surface area of

the unit sphere Sn−1. We have the following:

ωn−1 =2π

n2

Γ ( n2)

and

vn =ωn−1

n=

2πn2

nΓ ( n2)=

πn2

Γ ( n2+1)

.

The easy proofs are based on the formula in Appendix A.1. We have

(√π)n

=∫

Rne−|x|

2

dx = ωn−1

∫ ∞

0e−r2

rn−1 dr ,

by switching to polar coordinates. Now change variables t = r2 to obtain that

πn2 =

ωn−1

2

∫ ∞

0e−tt

n2−1 dt =

ωn−1

2Γ(

n2

).

This proves the formula for the surface area of the unit sphere in Rn.

To compute vn, write again using polar coordinates

vn = |B(0,1)|=∫

|x|≤11dx =

∫

Sn−1

∫ 1

0rn−1 dr dθ =

1

nωn−1 .

Here is another way to relate the volume to the surface area. Let B(0,R) be the

ball in Rn of radius R > 0 centered at the origin. Then the volume of the shell

B(0,R+ h) \B(0,R) divided by h tends to the surface area of B(0,R) as h→ 0. In

other words, the derivative of the volume of B(0,R) with respect to the radius R is

equal to the surface area of B(0,R). Since the volume of B(0,R) is vnRn, it follows

that the surface area of B(0,R) is nvnRn−1. Taking R = 1, we deduce ωn−1 = nvn.

A.4 Computation of Integrals Using Gamma Functions

Let k1, . . . ,kn be nonnegative even integers. The integral

∫

Rnx

k11 · · ·xkn

n e−|x|2

dx1 · · ·dxn =n

∏j=1

∫ +∞

−∞x

k j

j e−x2

j dx j =n

∏j=1

Γ(k j +1

2

)

expressed in polar coordinates is equal to

(∫

Sn−1θ k1

1 · · ·θ knn dθ

)∫ ∞

0rk1+···+knrn−1e−r2

dr ,


where θ = (θ1, . . . ,θn). This leads to the identity

∫

Sn−1θ k1

1 · · ·θ knn dθ = 2Γ

(k1 + · · ·+ kn +n

2

)−1 n

∏j=1

Γ(k j +1

2

).

Another classical integral that can be computed using gamma functions is the

following:∫ π/2

0(sinϕ)a(cosϕ)b dϕ =

1

2

Γ ( a+12)Γ ( b+1

2)

Γ ( a+b+22

),

whenever a and b are complex numbers with Rea >−1 and Reb >−1.

Indeed, change variables u = (sinϕ)2; then du = 2(sinϕ)(cosϕ)dϕ , and the pre-

ceding integral becomes

1

2

∫ 1

0u

a−12 (1−u)

b−12 du =

1

2B(a+1

2,

b+1

2

)=

1

2

Γ ( a+12)Γ ( b+1

2)

Γ ( a+b+22

).

A.5 Meromorphic Extensions of B(z,w) and Γ (z)

Using the identity Γ (z+ 1) = zΓ (z), we can easily define a meromorphic exten-

sion of the gamma function on the whole complex plane starting from its known

values on the right half-plane. We give an explicit description of the meromorphic

extension of Γ (z) on the whole plane. First write

Γ (z) =∫ 1

0tz−1e−tdt +

∫ ∞

1tz−1e−tdt

and observe that the second integral is an analytic function of z for all z ∈ C. Write

the first integral as

∫ 1

0tz−1

e−t −

N

∑j=0

(−t) j

j!

dt +

N

∑j=0

(−1) j/ j!

z+ j.

The last integral converges when Rez>−N−1, since the expression inside the curly

brackets is O(tN+1) as t → 0. It follows that the gamma function can be defined to be

an analytic function on Rez >−N−1 except at the points z =− j, j = 0,1, . . . ,N, at

which it has simple poles with residues(−1) j

j!. Since N was arbitrary, it follows that

the gamma function has a meromorphic extension on the whole plane.

In view of the identity

B(z,w) =Γ (z)Γ (w)

Γ (z+w),

the definition of B(z,w) can be extended to C×C. It follows that B(z,w) is a mero-

morphic function in each argument.

A.6 Asymptotics of Γ (x) as x→ ∞ 567

A.6 Asymptotics of Γ (x) as x→ ∞

We now derive Stirling’s formula:

limx→∞

Γ (x+1)(xe

)x√2πx

= 1 .

First change variables t = x+ sx

√2x

to obtain

Γ (x+1) =∫ ∞

0e−ttx dt =

(x

e

)x√2x

∫ +∞

−√

x/2

(1+ s

√2x

)x

e2s√

x/2ds .

Setting y =√

x2, we obtain

Γ (x+1)(xe

)x√2x

=∫ +∞

−∞

((1+ s

y

)y

es

)2y

χ(−y,∞)(s)ds.

To show that the last integral converges to√π as y→ ∞, we need the following:

(1) The fact that

limy→∞

((1+ s/y

)y

es

)2y

→ e−s2

,

which follows easily by taking logarithms and applying L’Hopital’s rule twice.

(2) The estimate, valid for y≥ 1,

((1+ s

y

)y

es

)2y

≤

⎧⎪⎨⎪⎩

(1+ s)2

eswhen s≥ 0,

e−s2when −y < s < 0,

which can be easily checked using calculus. Using these facts, the Lebesgue dom-

inated convergence theorem, the trivial fact that χ−y<s<∞ → 1 as y → ∞, and the

identity in Appendix A.1, we obtain that

limx→∞

Γ (x+1)(xe

)x√2x

= limy→∞

∫ +∞

−∞

((1+ s

y

)y

es

)2y

χ(−y,∞)(s)ds

=∫ +∞

−∞e−s2

ds

=√π.

As a consequence of Stirling’s formula, for any t > 0, we obtain

limx→∞

Γ (x)

Γ (x+ t)= 0 .

A.7 Euler’s Limit Formula for the Gamma Function

For n a positive integer and Rez > 0 we consider the functions

Γn(z) =∫ n

0

(1− t

n

)n

tz−1 dt

We show that

Γn(z) =n!nz

z(z+1) · · ·(z+n)

and we obtain Euler’s limit formula for the gamma function

limn→∞

Γn(z) = Γ (z) .

We write Γ (z)−Γn(z) = I1(z)+ I2(z)+ I3(z), where

I1(z) =∫ ∞

ne−ttz−1 dt ,

I2(z) =∫ n

n/2

(e−t −

(1− t

n

)n)

tz−1 dt ,

I3(z) =∫ n/2

0

(e−t −

(1− t

n

)n)

tz−1 dt .

Obviously I1(z) tends to zero as n→∞. For I2 and I3 we have that 0≤ t < n, and by

the Taylor expansion of the logarithm we obtain

log(

1− t

n

)n

= n log(

1− t

n

)=−t−L ,

where

L =t2

n

(1

2+

1

3

t

n+

1

4

t2

n2+ · · ·

).

It follows that

0 < e−t −(

1− t

n

)n

= e−t − e−Le−t ≤ e−t ,

and thus I2(z) tends to zero as n→ ∞. For I3 we have t/n≤ 1/2, which implies that

L≤ t2

n

∞

∑k=0

1

(k+1)2k−1=

t2

nc .

A.7 Euler’s Limit Formula for the Gamma Function 569

Consequently, for t/n≤ 1/2 we have

0≤ e−t −(

1− t

n

)n

= e−t(1− e−L)≤ e−tL≤ e−t ct2

n.

Plugging this estimate into I3, we deduce that

|I3(z)| ≤c

nΓ (Rez+2) ,

which certainly tends to zero as n→ ∞.

Next, n integrations by parts give

Γn(z) =n

nz

n−1

n(z+1)

n−2

n(z+2)· · · 1

n(z+n−1)

∫ n

0tz+n−1 dt =

n!nz

z(z+1) · · ·(z+n).

This can be written as

1 = Γn(z)zexp

z(

1+1

2+

1

3+ · · ·+ 1

n− logn

) n

∏k=1

(1+

z

k

)e−z/k .

Taking limits as n→ ∞, we obtain an infinite product form of Euler’s limit formula,

1 = Γ (z)zeγz∞

∏k=1

(1+

z

k

)e−z/k ,

where Rez > 0 and γ is Euler’s constant

γ = limn→∞

1+1

2+

1

3+ · · ·+ 1

n− logn .

The infinite product converges uniformly on compact subsets of the complex plane

that excludes z= 0,−1,−2, . . . , and thus it represents a holomorphic function in this

domain. This holomorphic function multiplied by Γ (z)zeγz is equal to 1 on Rez > 0

and by analytic continuation it must be equal to 1 on C\0,−1,−2, . . .. But Γ (z)has simple poles, while the infinite product vanishes to order one, at the nonpositive

integers. We conclude that Euler’s limit formula holds for all complex numbers z;

consequently, Γ (z) has no zeros and Γ (z)−1 is entire.

An immediate consequence of Euler’s limit formula is the identity

1

|Γ (x+ iy)|2 =1

|Γ (x)|2∞

∏k=0

(1+

y2

(k+ x)2

),

which holds for x and y real with x /∈ 0,−1,−2, . . .. As a consequence we have

that

|Γ (x+ iy)| ≤ |Γ (x)|


and also that1

|Γ (x+ iy)| ≤1

|Γ (x)|eC(x)|y|2 ,

where

C(x) =1

2

∞

∑k=0

1

(k+ x)2,

whenever y ∈ R and x ∈ R \ 0,−1,−2, . . .. Before we find a similar estimate for

x ∈ 0,−1,−2, . . . we provide a simpler expression for this estimate when x > 0.

When x > 0 we have

C(x)≤ 1

2x2+

1

2

∞

∑k=1

1

(k+ x)2≤ 1

2x2+

1

2

∫ ∞

0

dt

(t + x)2=

1

2x2+

1

2x

∫ ∞

1

dt

t2=

1

2x2+

1

2x.

Thus we conclude that when x > 0 and y ∈ R we have

1

|Γ (x+ iy)| ≤1

|Γ (x)|emaxx−2,x−1|y|2 .

When x = 0 we write Γ (iy)iy =Γ (1+ iy) and use the preceding inequality to obtain

1

|Γ (iy)| ≤|iy||Γ (1)|e

|y|2 = |y|e|y|2

and more generally for x =−N ∈ −1,−2, . . . and y ∈ R we obtain by induction

1

|Γ (−N + iy)| ≤ |iy| |1+ iy| |2+ iy| · · · |N + iy|e|y|2 .

A.8 Reflection and Duplication Formulas for the Gamma

Function

The reflection formula relates the values of the gamma function of a complex num-

ber z and its reflection about the point 1/2 in the following way:

sin(πz)

π=

1

Γ (z)

1

Γ (1− z).

The duplication formula relates the entire functions Γ (2z)−1 and Γ (z)−1 as follows:

1

Γ (z)Γ (z+ 12)=π−

12 22z−1

Γ (2z).

A.8 Reflection and Duplication Formulas for the Gamma Function 571

Both of these could be proved using Euler’s limit formula. The reflection formula

also uses the identity∞

∏k=1

(1− z2

k2

)=

sin(πz)

πz,

while the duplication formula makes use of the fact that

limn→∞

(n!)2 22n+1

(2n)!n1/2= 2π1/2 .

These and other facts related to the gamma function can be found in Olver [271].

Appendix B

Bessel Functions

B.1 Definition

We survey some basics from the theory of Bessel functions Jν of complex order

ν with Reν > −1/2. We define the Bessel function Jν of order ν by its Poisson

representation formula

Jν(t) =

(t2

)ν

Γ (ν+ 12)Γ ( 1

2)

∫ +1

−1eits(1− s2)ν

ds√1− s2

,

where Reν > −1/2 and t ≥ 0. Although this definition is also valid when t is a

complex number, for the applications we have in mind, it suffices to consider the

case that t is real and nonnegative; in this case Jν(t) is also a real number.

B.2 Some Basic Properties

Let us summarize a few properties of Bessel functions. We take t > 0.

(1) We have the following recurrence formula:

d

dt

(t−νJν(t)

)=−t−νJν+1(t), Reν >−1/2.

(2) We also have the companion recurrence formula:

d

dt

(tνJν(t)

)= tνJν−1(t), Reν > 1/2.



573

574 B Bessel Functions

(3) Jν(t) satisfies the differential equation:

t2 d2

dt2(Jν(t))+ t

d

dt(Jν(t))+(t2−ν2)Jν(t) = 0 .

(4) If ν ∈ Z+, then we have the following identity, which was taken by Bessel as the

definition of Jν for integer ν :

Jν(t) =1

2π

∫ 2π

0eit sinθ e−iνθ dθ =

1

2π

∫ 2π

0cos(t sinθ −νθ)dθ .

(5) For Reν >−1/2 we have the following identity:

Jν(t) =1

Γ ( 12)

( t

2

)ν ∞

∑j=0

(−1) j Γ ( j+ 12)

Γ ( j+ν+1)

t2 j

(2 j)!,

which can also be written as

Jν(t) =∞

∑j=0

(−1) j

j!

1

Γ ( j+ν+1)

(t

2

)2 j+ν

,

using the equality (2 j)! = 22 j j!( j− 12)( j− 3

2) · · · 1

2= 22 j j!Γ ( j+ 1

2)Γ ( 1

2)−1.

(6) For Reν > 1/2 the identity below is valid:

d

dt(Jν(t)) =

1

2

(Jν−1(t)− Jν+1(t)

).

We first verify property (1). We have

d

dt

(t−νJν(t)

)=

i

2νΓ (ν+ 12)Γ ( 1

2)

∫ 1

−1seits(1− s2)ν−

12 ds

=i

2νΓ (ν+ 12)Γ ( 1

2)

∫ 1

−1

it

2eits (1− s2)ν+

12

ν+ 12

ds

= − t−νJν+1(t),

where we integrated by parts and used the fact that Γ (x+1) = xΓ (x).We now prove Property (2) for Reν > 1/2:

d

dt

(tνJν(t)

)

=2νt2ν−12−ν

Γ (ν+ 12)Γ ( 1

2)

∫ 1

−1eits(1− s2)ν−

12 ds+

t2ν2−ν

Γ (ν+ 12)Γ ( 1

2)

i

∫ 1

−1eitsis(1− s2)ν−

12 ds

B.2 Some Basic Properties 575

=2νt2ν−12−ν

Γ (ν+ 12)Γ ( 1

2)

∫ 1


12 ds+

t2ν2−ν

Γ (ν+ 12)Γ ( 1

2)

∫ 1

−1

(eits

t

)′(1− s2)ν−

12 sds

=2νt2ν−12−ν

Γ (ν+ 12)Γ ( 1

2)

∫ 1


12 ds− t2ν2−ν

Γ (ν+ 12)Γ ( 1

2)

∫ 1

−1

eits

t

((1− s2)ν−

12 s)′

ds

=t2ν−12−ν

Γ (ν+ 12)Γ ( 1

2)

∫ 1

−1eits

[2ν(1− s2)ν−

12 −

((1− s2)ν−

12 s)′]

ds

=t2ν−12−ν

Γ (ν+ 12)Γ ( 1

2)

∫ 1

−1eits(2ν−1)(1− s2)ν−

32 ds

=t2ν−12−(ν−1)

Γ (ν− 12)Γ ( 1

2)

∫ 1


12

ds√1− s2

= tνJν−1(t) .

We proceed with the proof of property (3). A calculation using the definition of

the Bessel function gives that the left-hand side of (3) is equal to

2−ν tν+1

Γ (ν+ 12)Γ ( 1

2)

∫ +1

−1eist

((1− s2)t +2is(ν+ 1

2)

)(1− s2)ν−

12 ds ,

which in turn is equal to

−i2−ν tν+1

Γ (ν+ 12)Γ ( 1

2)

∫ +1

−1

d

ds

(eist(1− s2)ν+

12)

ds = 0 .

Property (4) can be derived directly from (1). Define

Gν(t) =1

2π

∫ 2π

0eit sinθ e−iνθ dθ ,

for ν = 0,1,2, . . . and t > 0. We can show easily that G0 = J0. If we had

d

dt

(t−νGν(t)

)=−t−νGν+1(t), t > 0,

for ν ∈ Z+, we would immediately conclude that Gν = Jν for ν ∈ Z+. We have

d

dt

(t−νGν(t)

)= − t−ν

(ν

tGν(t)−

dGν

dt(t)

)

= − t−ν∫ 2π

0

ν

2πteit sinθ e−iνθ − 1

2π

(d

dteit sinθ

)e−iνθ dθ

= − t−ν

2π

∫ 2π

0i

d

dθ

(eit sinθ−iνθ

t

)+(cosθ − isinθ)eit sinθ e−iνθ dθ

= − t−ν

2π

∫ 2π

0eit sinθ e−i(ν+1)θ dθ

= − t−νGν+1(t) .


For t real, the identity in (5) can be derived by inserting the expression

∞

∑j=0

(−1) j (ts)2 j

(2 j)!+ isin(ts)

for eits in the definition of the Bessel function Jν(t) in Appendix B.1. Algebraic

manipulations yield

Jν(t) =(t/2)ν

Γ ( 12)

∞

∑j=0

(−1) j 1

Γ (ν+ 12)

t2 j

(2 j)!2

∫ 1

0s2 j−1(1− s2)ν−

12 sds

=(t/2)ν

Γ ( 12)

∞

∑j=0

(−1) j 1

Γ (ν+ 12)

t2 j

(2 j)!

Γ ( j+ 12)Γ (ν+ 1

2)

Γ ( j+ν+1)

=(t/2)ν

Γ ( 12)

∞

∑j=0

(−1) j Γ ( j+ 12)

Γ ( j+ν+1)

t2 j

(2 j)!.

To derive property (6) we first multiply (1) by tν and (2) by t−ν ; then we use the

product rule for differentiation and we add the resulting expressions.

For further identities on Bessel functions, one may consult Watson’s mono-

graph [371].

B.3 An Interesting Identity

Let Reµ >− 12, Reν >−1, and t > 0. Then the following identity is valid:

∫ 1

0Jµ(ts)s

µ+1(1− s2)ν ds =Γ (ν+1)2ν

tν+1Jµ+ν+1(t) .

To prove this identity we use formula (5) in Appendix B.2. We have

∫ 1

0Jµ(ts)s

µ+1(1− s2)ν ds

=

(t2

)µ

Γ ( 12)

∫ 1

0

∞

∑j=0

(−1) jΓ ( j+ 12) t2 j

Γ ( j+µ+1)(2 j)!s2 j+µ+µ(1− s2)νsds

=1

2

(t2

)µ

Γ ( 12)

∞

∑j=0

(−1) jΓ ( j+ 12) t2 j

Γ ( j+µ+1)(2 j)!

∫ 1

0u j+µ(1−u)ν du

=1

2

(t2

)µ

Γ ( 12)

∞

∑j=0

(−1) jΓ ( j+ 12) t2 j

Γ ( j+µ+1)(2 j)!

Γ (µ+ j+1)Γ (ν+1)

Γ (µ+ν+ j+2)

=2νΓ (ν+1)

tν+1

(t2

)µ+ν+1

Γ ( 12)

∞

∑j=0

(−1) jΓ ( j+ 12) t2 j

Γ ( j+µ+ν+2)(2 j)!

=Γ (ν+1)2ν

tν+1Jµ+ν+1(t) .

B.5 The Fourier Transform of a Radial Function on Rn 577

B.4 The Fourier Transform of Surface Measure on Sn−1

Let dσ denote surface measure on Sn−1 for n≥ 2. Then the following is true:

dσ(ξ ) =∫

Sn−1e−2πiξ ·θdθ =

2π

|ξ | n−22

J n−22(2π|ξ |) .

To see this, use the result in Appendix D.3 to write

dσ(ξ ) =∫

Sn−1e−2πiξ ·θ dθ

=2π

n−12

Γ ( n−12)

∫ +1

−1e−2πi|ξ |s(1− s2)

n−22

ds√1− s2

=2π

n−12

Γ ( n−12)

Γ ( n−22

+ 12)Γ ( 1

2)

(π|ξ |) n−22

J n−22(2π|ξ |)

=2π

|ξ | n−22

J n−22(2π|ξ |) .

B.5 The Fourier Transform of a Radial Function on Rn

Let f (x) = f0(|x|) be a radial function defined on Rn, where f0 is defined on [0,∞).Then the Fourier transform of f is given by the formula

f (ξ ) =2π

|ξ | n−22

∫ ∞

0f0(r)J n

2−1(2πr|ξ |)r n2 dr .

To obtain this formula, use polar coordinates to write

f (ξ ) =∫

Rnf (x)e−2πiξ ·x dx

=∫ ∞

0

∫

Sn−1f0(r)e

−2πiξ ·rθdθ rn−1dr

=∫ ∞

0f0(r) dσ(rξ )rn−1dr

=∫ ∞

0f0(r)

2π

(r|ξ |) n−22

J n−22(2πr|ξ |)rn−1dr

=2π

|ξ | n−22

∫ ∞

0f0(r)J n

2−1(2πr|ξ |)r n2 dr .


As an application we take f (x) = χB(0,1)(x), where B(0,1) is the unit ball in Rn.

We obtain

(χB(0,1)) (ξ ) =2π

|ξ | n−22

∫ 1

0J n

2−1(2π|ξ |r)rn2 dr =

J n2(2π|ξ |)|ξ | n

2

,

in view of the result in Appendix B.3. More generally, for Reλ >−1, let

mλ (ξ ) =

(1−|ξ |2)λ for |ξ | ≤ 1,

0 for |ξ |> 1.

Then

mλ∨ (x) =

2π

|x| n−22

∫ 1

0J n

2−1(2π|x|r)rn2 (1− r2)λ dr =

Γ (λ +1)

πλ

J n2+λ

(2π|x|)|x| n

2+λ,

using again the identity in Appendix B.3.

B.6 Bessel Functions of Small Arguments

We seek the behavior of Jk(r) as r → 0+. We fix ν ∈ C with Reν > − 12. Then we

have the identity

Jν(r) =rν

2νΓ (ν+1)+Sν(r) ,

where

Sν(r) =(r/2)ν

Γ (ν+ 12)Γ ( 1

2)

∫ +1

−1(eirt −1)(1− t2)ν−

12 dt

and Sν satisfies

|Sν(r)| ≤2−ReνrReν+1

(Reν+1) |Γ (ν+ 12)|Γ ( 1

2).

To prove this estimate we note that

Jν(r) =(r/2)ν

Γ (ν+ 12)Γ ( 1

2)

∫ +1

−1(1− t2)ν−

12 dt +Sν(r)

=(r/2)ν

Γ (ν+ 12)Γ ( 1

2)

∫ π

0(sin2 φ)ν−

12 (sinφ)dφ +Sν(r)

=(r/2)ν

Γ (ν+ 12)Γ ( 1

2)

Γ (ν+ 12)Γ ( 1

2)

Γ (ν+1)+Sν(r) ,

where we evaluated the last integral using the result in Appendix A.4. Using that

|eirt −1| ≤ r|t|, we deduce the assertion regarding the size of |Sν(r)|.

B.7 Bessel Functions of Large Arguments 579

It follows from these facts and the estimate in Appendix A.7 that for 0 < r ≤ 1

and Reν >−1/2 we have

|Jν(r)| ≤C0(Reν)emax

(1

(Reν+ 12)2, 1

Reν+ 12

)|Imν |2

rReν ,

where C0 is a constant that depends smoothly on Reν ∈ (−1/2,∞).

B.7 Bessel Functions of Large Arguments

For r > 0 and complex numbers ν with Reν >−1/2 we prove the identity

Jν(r)=(r/2)ν

Γ (ν+ 12)Γ ( 1

2)

[ie−ir

∫ ∞

0e−rt(t2+2it)ν−

12 dt− ieir

∫ ∞

0e−rt(t2−2it)ν−

12 dt

].

Fix 0 < δ < 1/10 < 10 < R < ∞. We consider the region Ωδ ,R in the complex

plane whose boundary is the set consisting of the interval [−1+ δ ,1− δ ] union a

quarter circle centered at 1 of radius δ from 1−δ to 1+ iδ , union the line segments

from 1+ iδ to 1+ iR, from 1+ iR to −1+ iR, and from −1+ iR to −1+ iδ , union

a quarter circle centered at −1 of radius δ from −1+ iδ to −1+δ . This is a simply

connected region on the interior of which the holomorphic function (1− z2) has no

zeros. Since Ωδ ,R is contained in the complement of the negative imaginary axis,

there is a holomorphic branch of the logarithm such that log(t) is real, log(−t) =log |t|+ iπ , and log(it) = log |t|+ iπ/2 for t > 0. Since the function log(1− z2) is

well defined and holomorphic in Ωδ ,R, we may define the holomorphic function

(1− z2)ν−12 = e(ν−

12 ) log(1−z2)

for z ∈Ωδ ,R. Since eirz(1− z2)ν−12 has no poles in Ωδ ,R, Cauchy’s theorem yields

i

∫ R

δeir(1+it)(t2−2it)ν−

12 dt +

∫ 1−δ

−1+δeirt(1− t2)ν−

12 dt

+ i

∫ δ

Reir(−1+it)(t2 +2it)ν−

12 dt +E(δ ,R) = 0 ,

where E(δ ,R) is the sum of the integrals over the two small quarter-circles of radius

δ and the line segment from 1+ iR to −1+ iR. The first two of these integrals are

bounded by constants times δ , the latter by a constant times R2Reν−1e−rR; hence

E(δ ,R)→ 0 as δ → 0 and R→ ∞. We deduce the identity

∫ +1

−1eirt(1−t2)ν−

12 dt = ie−ir

∫ ∞

0e−rt(t2+2it)ν−

12 dt− ieir

∫ ∞

0e−rt(t2−2it)ν−

12 dt .


Estimating the two integrals on the right by putting absolute values inside and mul-

tiplying by the missing factor rν2−ν(Γ (ν+ 12)Γ ( 1

2))−1, we obtain

|Jν(r)| ≤ 2(r/2)Reν e

π2 |Imν |

|Γ (ν+ 12)|Γ ( 1

2)

∫ ∞

0e−rttReν− 1

2(√

t2 +4)Reν− 1

2 dt ,

since the absolute value of the argument of t2±2it is at most π/2. When Reν > 1/2,

we use the inequality (√

t2 +4)Reν− 12 ≤ 2Reν− 3

2

(tReν− 1

2 +2Reν− 12

)to get

|Jν(r)| ≤ 2(r/2)Reν e

π2 |Imν |

|Γ (ν+ 12)|Γ ( 1

2)

2Reν− 32

[Γ (2Reν)

r2Reν+2Reν Γ (Reν+ 1

2)

rReν+ 12

].

When 1/2≥ Reν >−1/2 we use that(√

t2 +4)Reν− 1

2 ≤ 1 to deduce that

|Jν(r)| ≤ 2(r/2)Reν e

π2 |Imν |

|Γ (ν+ 12)|Γ ( 1

2)

Γ (Reν+ 12)

rReν+ 12

.

These estimates yield that for Reν >−1/2 and r ≥ 1 we have

|Jν(r)| ≤C1(Reν) e

(max((Reν+ 1

2 )−2,(Reν+ 1

2 )−1)+ π

2

)|Imν |2

r−1/2

using the result in Appendix A.7, where C1 is a constant that depends smoothly on

Reν on the interval (−1/2,∞).

B.8 Asymptotics of Bessel Functions

We obtain asymptotics for Jν(r) as r → ∞ whenever Reν > −1/2. We have the

following identity for r > 0:

Jν(r) =

√2

πrcos

(r− πν

2− π

4

)+Rν(r) ,

where Rν is given by

Rν(r) =(2π)−

12 rν

Γ (ν+ 12)

ei(r− πν2 − π

4 )∫ ∞

0e−rttν+

12[(1+ it

2)ν−

12 −1

]dt

t

+(2π)−

12 rν

Γ (ν+ 12)

e−i(r− πν2 − π

4 )∫ ∞

0e−rttν+

12[(1− it

2)ν−

12 −1

]dt

t

and satisfies |Rν(r)| ≤Cν r−3/2 whenever r ≥ 1.

B.8 Asymptotics of Bessel Functions 581

To see the validity of this identity we write

ie−ir(t2 +2it)ν−12 = (2t)ν−

12 e−i(r− νπ

2 − π4 )(1− it

2)ν−

12 ,

−ieir(t2−2it)ν−12 = (2t)ν−

12 ei(r− νπ

2 − π4 )(1+ it

2)ν−

12 .

Inserting these expressions into the corresponding integrals in the formula proved in

Appendix B.7, adding and subtracting 1 from each term (1± it2)ν−

12 , and multiplying

by the missing factor (r/2)ν/Γ (ν+ 12)Γ ( 1

2), we obtain the claimed identity

Jν(r) =

√2

πrcos

(r− πν

2− π

4

)+Rν(r) .

It remains to estimate Rν(r). We begin by noting that for a,b real with a > −1

we have the pair of inequalities

|(1± iy)a+ib−1| ≤ 3(|a|+ |b|)(2

a+12 e

π2 |b|

)y when 0 < y < 1 ,

|(1± iy)a+ib−1| ≤ (1+ y2)a2 e

π2 |b|+1≤ 2

(2

a+12 e

π2 |b|

)ya when 1≤ y < ∞ .

The first inequality is proved by splitting into real and imaginary parts and applying

the mean value theorem in the real part. Taking ν− 12= a+ ib, y= t/2, and inserting

these estimates into the integrals appearing in Rν , we obtain

|Rν(r)| ≤2

12 Reν+ 5

4 eπ2 |Imν |rReν

(2π)1/2|Γ (ν+ 12)|

[3√

2|ν |2

∫ 2

0e−rttReν+ 3

2dt

t+

2√

2

2Reν

∫ ∞

2e−rtt2Reν dt

t

].

It follows that for all r > 0 we have

|Rν(r)| ≤ 22

12 Reνe

π2 |Imν |

|Γ (ν+ 12)|

[|ν | Γ (Reν+ 3

2)

r3/2+

r−Reν

2Reν

∫ ∞

2re−tt2Reν dt

t

]

≤ 22

12 Reνe

π2 |Imν |

|Γ (ν+ 12)|

[|ν | Γ (Reν+ 3

2)

r3/2+

2Reν

rReν

Γ (2Reν)

er

],

using that e−t ≤ e−t/2e−r for t ≥ 2r. We conclude that for r ≥ 1 and Reν > −1/2

we have

|Rν(r)| ≤2

12 Reνe

π2 |Imν |

Γ (Reν+ 12)

emax((Reν+ 12 )−2,(Reν+ 1

2 )−1)|Imν |2

[|ν | Γ (Reν+ 3

2)

r3/2+

2Reν

rReν

Γ (2Reν)

er

]

≤ C2(Reν)e

(max((Reν+ 1

2 )−2,(Reν+ 1

2 )−1)+ π

2 +1)|Imν |2

r−3/2 ,

via the result in Appendix A.7, where C2 is a constant that depends smoothly on

Reν on the interval (−1/2,∞).


B.9 Bessel Functions of general complex indices

We now discuss how to extend Jν(t) to complex values of ν when t > 0. Identity (5)

Jν(t) =∞

∑j=0

(−1) j

j!

1

Γ ( j+ν+1)

(t

2

)2 j+ν

of Appendix B.2 expresses Jν(t) as a power series when ν is a complex number with

Reν > −1/2. But this power series converges absolutely for all complex values of

ν by the ratio test. Indeed, for j sufficiently large we have

∣∣∣∣∣

(−1) j+1

( j+1)!1

Γ ( j+1+ν+1)

(t2

)2 j+2+ν

(−1) j

j!1

Γ ( j+ν+1)

(t2

)2 j+ν

∣∣∣∣∣=∣∣∣∣∣Γ ( j+ν+1)

(t2

)2

( j+1)Γ ( j+ν+2)

∣∣∣∣∣=∣∣∣∣

(t2

)2

( j+1)( j+ν+1)

∣∣∣∣

which tends to zero as j→ ∞, thus the ratio test applies.

Therefore the power series defines an entire function of ν . Since this entire func-

tion coincides with Jν on (− 12,∞)×R, by analytic continuation, Jν(t) has an entire

extension for all t > 0.

We obtain an integral representation of Jz(t) when −1/2 ≥ Rez > −3/2. Fix

t > 0. Then we write

Jz(t) =

(t2

)z

Γ (z+ 12)Γ ( 1

2)

∫ 1

−1eits(1− s2)z ds√

1− s2

=2(

t2

)z

Γ (z+ 12)Γ ( 1

2)

∫ 1

0cos(ts)(1− s2)z− 1

2 ds

=

(t2

)z

Γ (z+ 12)Γ ( 1

2)

∫ 1

0

cos(t√

1−u)√1−u

uz− 12 du ,

where the last step follows by the change of variables u = 1− s2. By the mean value

theorem, there is a constant c in (0,1) such that

(t2

)z

Γ (z+ 12)Γ ( 1

2)

∫ 1

0

cos(t√

1−u)√1−u

uz− 12 du

=

(t2

)z

Γ (z+ 12)√π

∫ 1

0

d

du

[cos(t

√1−u)√

1−u

](c) uz+ 1

2 du+

(t2

)zcos(t)√π

∫ 1

0uz− 1

2 du

Γ (z+ 12)

=

(t2

)z

Γ (z+ 12)√π

∫ 1

0

d

du

[cos(t

√1−u)√

1−u

](c) uz+ 1

2 du+

(t2

)zcos(t)√π

1

Γ (z+ 32),

B.9 Bessel Functions of general complex indices 583

where the second identity is due to the fact that the integral∫ 1

0 uz− 12 du is equal to

(z+ 12)−1 when Rez >−1/2 and thus when divided by Γ (z+ 1

2) can be analytically

extended to an entire function. It follows from these calculations that

J− 12(t) =

√2

π

cos(t)√t

.

We now estimate Jz(t) when −3/2 < Rez≤−1/2. Given q ∈ (0,1) we write

Jz(t) = J1z (t;q)+ J2

z (t;q)+ J3z (t;q) ,

where

J1z (t;q) =

(t2

)z

Γ (z+ 12)Γ ( 1

2)

∫ q

0

d

du

[cos(t

√1−u)√

1−u

](c) uz+ 1

2 du

J2z (t;q) =

(t2

)zcos(t)√π

qz+ 12

Γ (z+ 32)

J3z (t;q) =

(t2

)z

Γ (z+ 12)Γ ( 1

2)

∫ 1

q

cos(t√

1−u)√1−u

uz− 12 du

for some c satisfying 0 < c < q.

Suppose that t > 2. Then we pick q = 1/t. We have

|J1z (t;1/t)| ≤

(t2

)Rez

∣∣Γ (z+ 12)∣∣√π

t(√

1−1/t)3

1

Rez+ 32

(1

t

)Rez+ 32 ≤ C1(Rez) t−

12∣∣Γ (z+ 1

2)∣∣ ,

for some constant C1 depending on Rez. The second term obviously satisfies

|J2z (t;1/t)| ≤ C2(Rez) t−

12∣∣Γ (z+ 3

2)∣∣ .

In J3z (t;1/t) we write

cos(t√

1−u)√1−u

= −2t

ddu

[sin(t

√1−u)

]and integrate by parts to

obtain the estimate

|J3z (t;1/t)| ≤

(t2

)Rez

∣∣Γ (z+ 12)∣∣√π

[2

t+

2

t12+Rez

]≤ C3(Rez) t−

12∣∣Γ (z+ 1

2)∣∣

when −1/2≥ Rez >−3/2. Thus for these z’s and t > 2 we have

|Jz(t)| ≤C0(Rez) t−12

[1

|Γ (z+ 12)|+

1

|Γ (z+ 32)|

].

When 0 < t ≤ 2 we choose q = 1/2. Then each of |Jkz (t;1/2)|, k = 1,2,3 is

bounded by a multiple of tRez and we obtain the estimate

|Jz(t)| ≤C00(Rez) tRez

[1

|Γ (z+ 12)|+

1

|Γ (z+ 32)|

].

when −1/2≥ Rez >−3/2 and 0 < t ≤ 2.

In the special case Rez =−1/2 or z =−1/2+ iθ , θ ∈ R, we deduce

|J− 12+iθ (t)| ≤Ct−

12

[1

|Γ (1+ iθ)| +1

|Γ (iθ)|

]≤Ct−

12 (1+ |θ |)e|θ |

2

for some constant C independent of all parameters; see Appendix A.7 for the last

inequality.

Appendix C

Rademacher Functions

C.1 Definition of the Rademacher Functions

The Rademacher functions are defined on [0,1] as follows: r0(t) = 1; r1(t) = 1 for

0≤ t ≤ 1/2 and r1(t) =−1 for 1/2 < t ≤ 1; r2(t) = 1 for 0≤ t ≤ 1/4, r2(t) =−1

for 1/4 < t ≤ 1/2, r2(t) = 1 for 1/2 < t ≤ 3/4, and r2(t) = −1 for 3/4 < t ≤ 1;

and so on. According to this definition, we have that r j(t) = sgn(sin(2 jπt)) for j =0,1,2, . . . . It is easy to check that the r j’s are mutually independent random variables

on [0,1]. This means that for all functions f j we have

∫ 1

0

n

∏j=0

f j(r j(t))dt =n

∏j=0

∫ 1

0f j(r j(t))dt .

To see the validity of this identity, we write its right-hand side as

f0(1)n

∏j=1

∫ 1

0f j(r j(t))dt = f0(1)

n

∏j=1

f j(1)+ f j(−1)

2

=f0(1)

2n ∑S⊂1,2,...,n

∏j∈S

f j(1)∏j/∈S

f j(−1)

and we observe that there is a one-to-one and onto correspondence between sub-

sets S of 1,2, . . . ,n and intervals Ik =[

k2n ,

k+12n

], k = 0,1, . . . ,2n−1, such that the

restriction of the function ∏nj=1 f j(r j(t)) on Ik is equal to

∏j∈S

f j(1)∏j/∈S

f j(−1) .

It follows that the last of the three equal displayed expressions is

f0(1)2n−1

∑k=0

∫

Ik

n

∏j=1

f j(r j(t))dt =∫ 1

0

n

∏j=0

f j(r j(t))dt .



585

586 C Rademacher Functions

C.2 Khintchine’s Inequalities

The following property of the Rademacher functions is of fundamental importance

and with far-reaching consequences in analysis:

For any 0 < p <∞ and for any real-valued square summable sequences a j and

b j we have

Bp

(∑

j

|a j + ib j|2)1

2

≤∥∥∥∑

j

(a j + ib j)r j

∥∥∥Lp([0,1])

≤ Ap

(∑

j

|a j + ib j|2)1

2

for some constants 0 < Ap,Bp < ∞ that depend only on p.

These inequalities reflect the orthogonality of the Rademacher functions in Lp

(especially when p = 2). Khintchine [193] was the first to prove a special form of

this inequality, and he used it to estimate the asymptotic behavior of certain random

walks. Later this inequality was systematically studied almost simultaneously by Lit-

tlewood [226] and by Paley and Zygmund [274], who proved the more general form

stated previously. The foregoing inequalities are usually referred to by Khintchine’s

name.

C.3 Derivation of Khintchine’s Inequalities

Both assertions in Appendix C.2 can be derived from an exponentially decaying

distributional inequality for the function

F(t) =∑j

(a j + ib j)r j(t) , t ∈ [0,1],

when a j, b j are square summable real numbers.

We first obtain a distributional inequality for the above function F under the

following three assumptions:

(a)The sequence b j is identically zero.

(b)All but finitely many terms of the sequence a j are zero.

(c)The sequence a j satisfies (∑ j |a j|2)1/2 = 1.

Let ρ > 0. Under assumptions (a), (b), and (c), independence gives

∫ 1

0eρ∑a jr j(t) dt =∏

j

∫ 1

0eρa jr j(t) dt

=∏j

eρa j + e−ρa j

2

≤∏j

e12ρ

2a2j = e

12ρ

2∑a2j = e

12ρ

2

,

C.3 Derivation of Khintchine’s Inequalities 587

where we used the inequality 12(ex+e−x)≤ e

12 x2

for all real x, which can be checked

using power series expansions. Since the same argument is also valid for−∑a jr j(t),we obtain that ∫ 1

0eρ |F(t)| dt ≤ 2e

12ρ

2

.

From this it follows that

eρα |t ∈ [0,1] : |F(t)|> α| ≤∫ 1

0eρ |F(t)| dt ≤ 2e

12ρ

2

and hence we obtain the distributional inequality

dF(α) = |t ∈ [0,1] : |F(t)|> α| ≤ 2e12ρ

2−ρα = 2e−12α

2

,

by picking ρ = α . The Lp norm of F can now be computed easily. Formula (1.1.6)

gives

∥∥F∥∥p

Lp =∫ ∞

0pα p−1dF(α)dα ≤

∫ ∞

0pα p−12e−

α2

2 dα = 2p2 pΓ (p/2) .

We have now proved that

∥∥F∥∥

Lp ≤√

2(

pΓ (p/2)) 1

p∥∥F

∥∥L2

under assumptions (a), (b), and (c).

We now dispose of assumptions (a), (b), and (c). Assumption (b) can be easily

eliminated by a limiting argument and (c) by a scaling argument. To dispose of

assumption (a), let a j and b j be real numbers. We have

∥∥∥∑j

(a j + ib j)r j

∥∥∥Lp≤

∥∥∥∣∣∑

j

a jr j

∣∣+∣∣∑

j

b jr j

∣∣∥∥∥

Lp

≤∥∥∥∑

j

a jr j

∥∥∥Lp+

∥∥∥∑j

b jr j

∥∥∥Lp

≤√

2(

pΓ (p/2)) 1

p

((∑

j

|a j|2)1

2+

(∑

j

|b j|2)1

2

)

≤√

2(

pΓ (p/2)) 1

p√

2(∑

j

|a j + ib j|2)1

2.

Let us now set Ap = 2(

pΓ (p/2))1/p

when p > 2. Since we have the trivial esti-

mate∥∥F

∥∥Lp ≤

∥∥F∥∥

L2 when 0 < p ≤ 2, we obtain the required inequality∥∥F

∥∥Lp ≤

Ap

∥∥F∥∥

L2 with

Ap =

1 when 0 < p≤ 2,

2 p1p Γ (p/2)

1p when 2 < p < ∞.

Using Sterling’s formula in Appendix A.6, we see that Ap is asymptotic to√

p as

p→ ∞.

We now discuss the converse inequality Bp

∥∥F∥∥

L2 ≤∥∥F

∥∥Lp . It is clear that∥∥F

∥∥L2 ≤

∥∥F∥∥

Lp when p ≥ 2 and we may therefore take Bp = 1 for p ≥ 2. Let us

now consider the case 0 < p < 2. Pick an s such that 2 < s < ∞. Find a 0 < θ < 1

such that1

2=

1−θp

+θ

s.

Then ∥∥F∥∥

L2 ≤∥∥F

∥∥1−θLp

∥∥F∥∥θ

Ls ≤∥∥F

∥∥1−θLp Aθs

∥∥F∥∥θ

L2 .

It follows that ∥∥F∥∥

L2 ≤ Aθ

1−θs

∥∥F∥∥

Lp .

We have now proved the inequality Bp

∥∥F∥∥

L2 ≤∥∥F

∥∥Lp with

Bp =

⎧⎪⎪⎨⎪⎪⎩

1 when 2≤ p < ∞,

sups>2

A

−1p− 1

212− 1

ss when 0 < p < 2.

Observe that the function s → A−(

1p− 1

2

)/(

12− 1

s

)s tends to zero as s → 2+ and as

s→ ∞. Hence it must attain its maximum for some s = s(p) in the interval (2,∞).We see that Bp ≥ 16 ·256−1/p when p < 2 by taking s = 4.

It is worthwhile to mention that the best possible values of the constants Ap and

Bp in Khintchine’s inequality are known when b j = 0. In this case Szarek [353]

showed that the best possible value of B1 is 1/√

2, and later Haagerup [141] found

that when b j = 0 the best possible values of Ap and Bp are the numbers

Ap =

1 when 0 < p≤ 2,

212 π−

12p Γ ( p+1

2) when 2 < p < ∞,

and

Bp =

⎧⎪⎪⎨⎪⎪⎩

212− 1

p when 0 < p≤ p0,

212 π−

12p Γ ( p+1

2) when p0 < p < 2,

1 when 2 < p < ∞,

where p0 = 1.84742 . . . is the unique solution of the equation 2Γ ( p+12) =

√π in the

interval (1,2).

C.5 Extension to Several Variables 589

C.4 Khintchine’s Inequalities for Weak Type Spaces

We note that the following weak type estimates are valid:

4− 1

p B p2

(∑

j

|a j + ib j|2)1

2

≤∥∥∥∑

j

(a j + ib j)r j

∥∥∥Lp,∞

≤ Ap

(∑

j

|a j + ib j|2)1

2

for all 0 < p < ∞.

Indeed, the upper estimate is a simple consequence of the fact that Lp is a sub-

space of Lp,∞. For the converse inequality we use the fact that Lp,∞([0,1]) is con-

tained in Lp/2([0,1]) and we have (see Exercise 1.1.11)

∥∥F∥∥

Lp/2 ≤ 41p∥∥F

∥∥Lp,∞ .

Since any Lorentz space Lp,q([0,1]) can be sandwiched between L2p([0,1]) and

Lp/2([0,1]), similar inequalities hold for all Lorentz spaces Lp,q([0,1]), 0 < p < ∞,

0 < q≤ ∞.

C.5 Extension to Several Variables

We first extend the inequality on the right in Appendix C.2 to several variables. For

a positive integer n we let

Fn(t1, . . . , tn) =∑j1

· · ·∑jn

c j1,..., jnr j1(t1) · · ·r jn(tn),

for t j ∈ [0,1], where c j1,..., jn is a sequence of complex numbers and Fn is a function

defined on [0,1]n.

For any 0 < p < ∞ and for any complex-valued square summable sequence of n

variables c j1,..., jn j1,..., jn , we have the following inequalities for Fn:

Bnp

(∑j1

· · ·∑jn

|c j1,..., jn |2)1

2

≤∥∥Fn

∥∥Lp([0,1]n)

≤ Anp

(∑j1

· · ·∑jn

|c j1,..., jn |2)1

2

,

where Ap,Bp are the constants in Appendix C.2. The norms are over [0,1]n.

The case n = 2 is indicative of the general case. For p≥ 2 we have

∫ 1

0

∫ 1

0|F2(t1, t2)|p dt1 dt2 ≤ Ap

p

∫ 1

0

(∑j1

∣∣∑j2

c j1, j2r j2(t2)∣∣2)p

2

dt2

≤ App

(∑j1

(∫ 1

0

∣∣∑j2

c j1, j2r j2(t2)∣∣p

dt2

)2p)p

2

≤ A2pp

(∑j1

∑j2

|c j1, jn |2)p

2

,

where we used Minkowski’s integral inequality (with exponent p/2≥ 1) in the sec-

ond inequality and the result in the case n = 1 twice.

The case p < 2 follows trivially from Holder’s inequality with constant Ap = 1.

The reverse inequalities follow exactly as in the case of one variable. Replacing Ap

by Anp in the argument, giving the reverse inequality in the case n = 1, we obtain the

constant Bnp.

Likewise one may extend the weak type inequalities of Appendix C.3 in several

variables.

Appendix D

Spherical Coordinates

D.1 Spherical Coordinate Formula

Switching integration from spherical coordinates to Cartesian is achieved via the

following identity:

∫

RSn−1

f (x)dσ(x) =∫ π

ϕ1=0· · ·

∫ π

ϕn−2=0

∫ 2π

ϕn−1=0f (x(ϕ))J(n,R,ϕ)dϕn−1 · · ·dϕ1,

where

x1 = Rcosϕ1 ,

x2 = Rsinϕ1 cosϕ2 ,

x3 = Rsinϕ1 sinϕ2 cosϕ3 ,

. . .

xn−1 = Rsinϕ1 sinϕ2 sinϕ3 · · ·sinϕn−2 cosϕn−1 ,

xn = Rsinϕ1 sinϕ2 sinϕ3 · · ·sinϕn−2 sinϕn−1 ,

and 0≤ ϕ1, . . . ,ϕn−2 ≤ π , 0≤ ϕn−1 = θ < 2π ,

x(ϕ) = (x1(ϕ1, . . . ,ϕn−1), . . . ,xn(ϕ1, . . . ,ϕn−1)) ,

and

J(n,R,ϕ) =

(n

∑k=1

∣∣∣det( ∂xi

∂ϕ j

)1≤i =k≤n1≤ j≤n−1

∣∣∣2) 1

2

= Rn−1(sinϕ1)n−2 · · ·(sinϕn−3)

2(sinϕn−2)

is the Jacobian of the transformation.



591

592 D Spherical Coordinates

D.2 A Useful Change of Variables Formula

The following formula is useful in computing integrals over the sphere Sn−1 when

n≥ 2. Let f be a function defined on Sn−1. Then we have

∫

RSn−1f (x)dσ(x) =

∫ +R

−R

∫

√R2−s2 Sn−2

f(s,θ

)dθ

Rds√R2− s2

.

To prove this formula, let ϕ ′ = (ϕ2, . . . ,ϕn−1) and

x′ = x′(ϕ ′) = (cosϕ2,sinϕ2 cosϕ3, . . . ,sinϕ2 · · ·sinϕn−2 sinϕn−1) .

Using the change of variables in Appendix D.1 we express

∫

RSn−1f (x)dσ(x)

as the iterated integral

∫ π

ϕ1=0

[∫ π

ϕ2=0· · ·

∫ 2π

ϕn−1=0f (Rcosϕ1,Rsinϕ1 x′(ϕ ′))J(n−1,1,ϕ ′)dϕ ′

]Rdϕ1

(Rsinϕ1)2−n,

and we can realize the expression inside the square brackets as

∫

Sn−2f (Rcosϕ1,Rsinϕ1 x′)dσ(x′) .

Consequently,

∫


∫ π

ϕ1=0

∫

Sn−2f (Rcosϕ1,Rsinϕ1 x′)dσ(x′)Rn−1(sinϕ1)

n−2dϕ1 ,

and the change of variables

s = Rcosϕ1 , ϕ1 ∈ (0,π),

ds =−Rsinϕ1 dϕ1 ,√

R2− s2 = Rsinϕ1 ,

yields

∫


∫ R

−R

∫

Sn−2f (s,

√R2− s2 θ)dθ

(√R2− s2

)n−2 Rds√R2− s2

.

Rescaling the sphere Sn−2 to√

R2− s2 Sn−2 yields the claimed identity.

D.4 The Computation of Another Integral over the Sphere 593

D.3 Computation of an Integral over the Sphere

Let K be a function on the line. We use the result in Appendix D.2 to show that for

n≥ 2 we have

∫

Sn−1K(x ·θ)dθ =

2πn−1

2

Γ(

n−12

)∫ +1

−1K(s|x|)

(√1− s2

)n−3ds

when x ∈ Rn \ 0. Let x′ = x/|x| and pick a matrix A ∈ O(n) such that Ae1 = x′,where e1 = (1,0, . . . ,0). We have

∫

Sn−1K(x ·θ)dθ =

∫

Sn−1K(|x|(x′ ·θ))dθ

=∫

Sn−1K(|x|(Ae1 ·θ))dθ

=∫

Sn−1K(|x|(e1 ·A−1θ))dθ

=∫

Sn−1K(|x|θ1)dθ

=

∫ +1

−1K(|x|s)ωn−2

(√1− s2

)n−2 ds√1− s2

= ωn−2

∫ +1

−1K(s|x|)

(√1− s2

)n−3ds ,

where ωn−2 = 2πn−1

2 Γ(

n−12

)−1is the surface area of Sn−2.

For example, we have

∫

Sn−1

dθ

|ξ ·θ |α = ωn−2

∫ +1

−1

1

|s|α |ξ |α (1− s2)n−3

2 ds =1

|ξ |α2π

n−12 Γ

(1−α

2

)

Γ(

n−α2

) ,

and the integral converges only when Reα < 1.

D.4 The Computation of Another Integral over the Sphere

We compute the following integral for n≥ 2:

∫

Sn−1

dθ

|θ − e1|α,

594 D Spherical Coordinates

where e1 = (1,0, . . . ,0). Applying the formula in Appendix D.2, we obtain

∫

Sn−1

dθ

|θ − e1|α=

∫ +1

−1

∫

θ∈√

1−s2 Sn−2

dθ

(|s−1|2 + |θ |2) α2ds√

1− s2

=∫ +1

−1ωn−2

(1− s2)n−2

2

((1− s)2 +1− s2

) α2

ds√1− s2

=ωn−2

2α2

∫ +1

−1

(1− s2)n−3

2

(1− s)α2

ds

=ωn−2

2α2

∫ +1

−1(1− s)

n−3−α2 (1+ s)

n−32 ds ,

which converges exactly when Reα < n−1.

D.5 Integration over a General Surface

Suppose that S is a hypersurface in Rn of the form S = (u,Φ(u)) : u ∈ D, where

D is an open subset of Rn−1 and Φ is a continuously differentiable mapping from D

to R. Let σ be the canonical surface measure on S. If g is a function on S, then we

have ∫

Sg(y)dσ(y) =

∫

Dg(x,Φ(x))

(1+

n−1

∑j=1

|∂ jΦ(x)|2)1

2dx .

Specializing to the sphere, we obtain

∫

Sn−1g(θ)dθ =

∫

ξ ′∈Rn−1

|ξ ′|<1

[g(ξ ′,

√1−|ξ ′|2)+g(ξ ′,−

√1−|ξ ′|2)

] dξ ′√1−|ξ ′|2

.

D.6 The Stereographic Projection

Define a map Π : Rn → Sn by the formula

Π(x1, . . . ,xn) =

(2x1

1+ |x|2 , . . . ,2xn

1+ |x|2 ,|x|2−1

1+ |x|2).

It is easy to see that Π is a one-to-one map from Rn onto the sphere Sn minus the

north pole en+1 = (0, . . . ,0,1). Its inverse is given by the formula

Π−1(θ1, . . . ,θn+1) =

(θ1

1−θn+1, . . . ,

θn

1−θn+1

).

D.6 The Stereographic Projection 595

The Jacobian of the map is verified to be JΠ (x) =(

21+|x|2

)−n, and the following

change of variables formulas are valid:

∫

SnF(θ)dθ =

∫

RnF(Π(x))JΠ (x)dx

and ∫

RnF(x)dx =

∫

SnF(Π−1(θ))JΠ−1(θ)dθ ,

where

JΠ−1(θ) =1

JΠ (Π−1(θ))=

(|θ1|2 + · · ·+ |θn|2 + |1−θn+1|2

2|1−θn+1|2

)n

.

Another interesting formula about the stereographic projection Π is

|Π(x)−Π(y)|= 2|x− y|(1+ |x|2)−1/2(1+ |y|2)−1/2 , x,y ∈ Rn .

Appendix E

Some Trigonometric Identities and Inequalities

The following inequalities are valid for t real:

0 < t <π

2=⇒ sin(t)< t < tan(t) ,

0 < |t|< π

2=⇒ 2

π<

sin(t)

t< 1 ,

−∞< t <+∞ =⇒ |sin(t)| ≤ |t| ,

−∞< t <+∞ =⇒ |1− cos(t)| ≤ |t|2

2,

−∞< t <+∞ =⇒ |1− eit | ≤ |t| ,

|t| ≤ π

2=⇒ |sin(t)| ≥ 2|t|

π,

|t| ≤ π =⇒ |1− cos(t)| ≥ 2|t|2π2

,

|t| ≤ π =⇒ |1− eit | ≥ 2|t|π

.

The following sum to product formulas are valid:

sin(a)+ sin(b) = 2 sin(a+b

2

)cos

(a−b

2

),

sin(a)− sin(b) = 2 cos(a+b

2

)sin

(a−b

2

),

cos(a)+ cos(b) = 2 cos(a+b

2

)cos

(a−b

2

),

cos(a)− cos(b) = −2 sin(a+b

2

)sin

(a−b

2

).



597

598 E Some Trigonometric Identities and Inequalities

The following identities are also easily proved:

N

∑k=1

cos(kx) = − 1

2+

sin((N + 12)x)

2sin( x2)

,

N

∑k=1

sin(kx) =cos( x

2)− cos((N + 1

2)x)

2sin( x2)

.

Appendix F

Summation by Parts

Let ak∞k=0, bk∞k=0 be two sequences of complex numbers. Then for N ≥ 1 we

haveN

∑k=0

akbk = ANbN −N−1

∑k=0

Ak(bk+1−bk),

where

Ak =k

∑j=0

a j .

More generally we have

N

∑k=M

akbk = ANbN −AM−1bM−N−1

∑k=M

Ak(bk+1−bk) ,

whenever 0≤M ≤ N, where A−1 = 0 and

Ak =k

∑j=0

a j

for k ≥ 0.



599

Appendix G

Basic Functional Analysis

A quasi-norm is a nonnegative functional ‖ · ‖ on a vector space X that satisfies

‖x + y‖X ≤ K(‖x‖X + ‖y‖X ) for some K ≥ 0 and all x,y ∈ X and also ‖λx‖X =|λ |‖x‖X for all scalars λ . When K = 1, the quasi-norm is called a norm. A quasi-

Banach space is a quasi-normed space that is complete with respect to the topology

generated by the quasi-norm. The proofs of the following theorems can be found in

several books including Albiac and Kalton [1], Kalton Peck and Roberts [188], and

Rudin [306].

The Hahn–Banach theorem. Let X be a normed vector space (over the real or

complex numbers), let Y be a subspace of X , and let P be a positively homogeneous

subadditive1 functional on X . Then for every linear functional Λ on Y that satisfies

|Λ(y)| ≤ P(y)

for all y ∈ Y , there is a linear functional Λ ′ on X such that

Λ ′(y) = Λ(y) for all y ∈ Y ,

|Λ ′(x)| ≤ P(x) for all x ∈ X .

In particular, every bounded linear functional on a subspace has an extension on

the entire space with the same norm.

Banach–Alaoglou theorem. Let X be a normed vector space and let X∗ be the

space of all bounded linear functionals on X . Then the closed unit ball of X∗ is

compact in the weak∗ topology.

A special case is the sequential version of this theorem, which asserts that the

closed unit ball of the dual space of a separable normed vector space is sequentially

compact in the weak∗ topology. Indeed, the weak∗ topology on the closed unit ball

1 this means that P(x) ≥ 0 for all x ∈ X , P(λx) = λP(x) for all λ > 0 and all x ∈ X , and that

P(x+ z)≤ P(x)+P(z) for all x,z ∈ X .



601

602 G Basic Functional Analysis

of the dual of a separable space is metrizable, and thus compactness and sequential

compactness are equivalent.

Open mapping theorem. Suppose that X and Y are quasi-Banach spaces and

T is a bounded surjective linear map from X onto Y . Then T maps open sets to

open sets, i.e., it is an open mapping. Moreover, if T is injective, then there exists a

constant K < ∞ such that for all x ∈ X we have

‖x‖X ≤ K‖T (x)‖Y .

Closed graph theorem. Suppose that X and Y are quasi-Banach spaces and T is

a linear map from X to Y whose graph is a closed set, i.e., whenever xk,x ∈ X and

(xk,T (xk)) → (x,y) in X ×Y for some y ∈ Y , then T (x) = y. Then T is a bounded

linear map from X to Y .

Uniform boundedness principle. Suppose that X is a quasi-Banach space, Y is

a quasi-normed space and (Tα)α∈I is a family of bounded linear maps from X to Y

such that for all x ∈ X there exists a Cx < ∞ such that

supα∈I

‖Tα(x)‖Y ≤Cx .

Then there exists a constant K < ∞ such that

supα∈I

‖Tα‖X→Y ≤ K .

Appendix H

The Minimax Lemma

Minimax type results are used in the theory of games and have their origin in the

work of Von Neumann [369]. Much of the theory in this subject is based on convex

analysis techniques. For instance, this is the case with the next proposition, which

is needed in the “difficult” inequality in the proof of the minimax lemma. We refer

to Fan [111] for a general account of minimax results. The following exposition is

based on the simple presentation in Appendix A2 of [122].

Minimax Lemma. Let A, B be convex subsets of certain vector spaces. Assume that

a topology is defined in B for which it is a compact Hausdorff space and assume that

there is a function Φ : A×B→ R⋃+∞ that satisfies the following:

(a) Φ( . ,b) is a concave function on A for each b ∈ B,

(b) Φ(a, .) is a convex function on B for each a ∈ A,

(c) Φ(a, .) is lower semicontinuous on B for each a ∈ A.

Then the following identity holds:

minb∈B

supa∈A

Φ(a,b) = supa∈A

minb∈B

Φ(a,b) .

To prove the lemma we need the following proposition:

Proposition. Let B be a convex compact subset of a vector space and suppose that

g j : B→R⋃+∞, j = 1,2, . . . ,n, are convex and lower semicontinuous functions.

If

max1≤ j≤n

g j(b)> 0 for all b ∈ B ,

then there exist nonnegative numbers λ1,λ2, . . . ,λn such that

λ1g1(b)+λ2g2(b)+ · · ·+λngn(b)> 0 for all b ∈ B .

Proof. We first consider the case n = 2. Define subsets of B

B1 = b ∈ B : g1(b)≤ 0, B2 = b ∈ B : g2(b)≤ 0 .



603

604 H The Minimax Lemma

If B1 = /0, we take λ1 = 1 and λ2 = 0, and we similarly deal with the case B2 = /0. If

B1 and B2 are nonempty, then they are closed and thus compact. The hypothesis of

the proposition implies that g2(b)> 0≥ g1(b) for all b∈ B1. Therefore, the function

−g1(b)/g2(b) is well defined and upper semicontinuous on B1 and thus attains its

maximum. The same is true for −g2(b)/g1(b) defined on B2. We set

µ1 = maxb∈B1

−g1(b)

g2(b)≥ 0 , µ2 = max

b∈B2

−g2(b)

g1(b)≥ 0 .

We need to find λ > 0 such that λg1(b)+ g2(b) > 0 for all b ∈ B. This is clearly

satisfied if b ∈ B1

⋃B2, while for b1 ∈ B1 and b2 ∈ B2 we have

λg1(b1)+g2(b1) ≥ (1−λµ1)g2(b1) ,

λg1(b2)+g2(b2) ≥ (λ −µ2)g1(b2) .

Therefore, it suffices to find a λ > 0 such that 1−λµ1 > 0 and λ −µ2 > 0. Such a

λ exists if and only if µ1µ2 < 1. To prove that µ1µ2 < 1, we can assume that µ1 = 0

and µ2 = 0. Then we take b1 ∈ B1 and b2 ∈ B2, for which the maxima µ1 and µ2 are

attained, respectively. Then we have

g1(b1)+µ1g2(b1) = 0 ,

g1(b2)+1

µ2g2(b2) = 0 .

But g1(b1)< 0 < g1(b2); thus taking bθ = θb1 +(1−θ)b2 for some θ in (0,1), we

have

g1(bθ )≤ θg1(b1)+(1−θ)g1(b2) = 0 .

Considering the same convex combination of the last displayed equations and using

this identity, we obtain that

µ1µ2θg2(b1)+(1−θ)g2(b2) = 0 .

The hypothesis of the proposition implies that g2(bθ )> 0 and the convexity of g2:

θg2(b1)+(1−θ)g2(b2)> 0 .

Since g2(b1) > 0, we must have µ1µ2g2(b1) < g2(b1), which gives µ1µ2 < 1. This

proves the required claim and completes the case n = 2.

We now use induction to prove the proposition for arbitrary n. Assume that the

result has been proved for n−1 functions. Consider the subset of B

Bn = b ∈ B : gn(b)≤ 0 .

If Bn = /0, we choose λ1 = λ2 = · · ·= λn−1 = 0 and λn = 1. If Bn is not empty, then it

is compact and convex and we can restrict g1,g2, . . . ,gn−1 to Bn. Using the induction

hypothesis, we can find λ1,λ2, . . . ,λn−1 ≥ 0 such that

H The Minimax Lemma 605

g0(b) = λ1g1(b)+λ2g2(b)+ · · ·+λn−1gn−1(b)> 0

for all b ∈ Bn. Then g0 and gn are convex lower semicontinuous functions on B, and

max(g0(b),gn(b)) > 0 for all b ∈ B. Using the case n = 2, which was first proved,

we can find λ0,λn ≥ 0 such that for all b ∈ B we have

0 < λ0g0(b)+λngn(b)

= λ0λ1g1(b)+λ0λ2g2(b)+ · · ·+λ0λn−1gn−1(b)+λngn(b).

This establishes the case of n functions and concludes the proof of the induction and

hence of the proposition. We now turn to the proof of the minimax lemma.

Proof. The fact that the left-hand side in the required conclusion of the minimax

lemma is at least as big as the right-hand side is obvious. We can therefore concen-

trate on the converse inequality. In doing this we may assume that the right-hand side

is finite. Without loss of generality we can subtract a finite constant from Φ(a,b),and so we can also assume that

supa∈A

minb∈B

Φ(a,b) = 0 .

Then, by hypothesis (c) of the minimax lemma, the subsets

Ba = b ∈ B : Φ(a,b)≤ 0, a ∈ A

of B are closed and nonempty, and we show that they satisfy the finite intersection

property. Indeed, suppose that

Ba1∩Ba2

∩·· ·∩Ban = /0

for some a1,a2, . . . ,an ∈ A. We write g j(b) = Φ(a j,b), j = 1,2, . . . ,n, and we ob-

serve that the conditions of the previous proposition are satisfied. Therefore we can

find λ1,λ2, . . . ,λn ≥ 0 such that for all b ∈ B we have

λ1Φ(a1,b)+λ2Φ(a2,b)+ · · ·+λnΦ(an,b)> 0 .

For simplicity we normalize the λ j’s by setting λ1 + λ2 + · · ·+ λn = 1. If we set

a0 = λ1a1 +λ2a2 + · · ·+λnan, the concavity hypothesis (a) gives

Φ(a0,b)> 0

for all b ∈ B, contradicting the fact that supa∈A minb∈BΦ(a,b) = 0. Therefore, the

family of closed subsets Baa∈A of B satisfies the finite intersection property. The

compactness of B now implies⋂

a∈A Ba = /0. Take b0 ∈⋂

a∈A Ba. Then Φ(a,b0)≤ 0

for every a ∈ A, and therefore

minb∈B

supa∈A

Φ(a,b)≤ supa∈A

Φ(a,b0)≤ 0

as required.

Appendix I

Taylor’s and Mean Value Theorem in SeveralVariables

I.1 Mutlivariable Taylor’s Theorem

For a multiindex α = (α1, . . . ,αn) ∈ (Z+∪0)n, we denote by |α|= α1 + · · ·+αn

its size, we define α! = α1! · · ·αn! its factorial, and we set

hα = hα11 · · ·hαn

n ,

where h = (h1, . . . ,hn); here 00 = 1.

Let k ∈ Z+∪0. Suppose a real-valued C k+1 function f is defined on an open

convex subset Ω of Rn. Suppose that x ∈Ω and x+h ∈Ω . Then we have the Taylor

expansion formula

f (x+h) = ∑|α |≤k

∂α f (x)

α!hα +R(h,x,k) ,

where the remainder R(h,x,k) can be expressed either in Langrange’s mean value

form

R(h,x,k) = ∑|α |=k+1

∂α f (x+ ch)

α!hα

for some c ∈ (0,1), or in integral form

R(h,x,k) = (k+1) ∑|α |=k+1

hα

α!

∫ 1

0(1− t)k∂α f (x+ th)dt .



607

608 I Taylor’s and Mean Value Theorem in Several Variables

I.2 The Mean value Theorem

Suppose that f is as above and k = 0. Then for given x,y ∈Ω we have

f (y)− f (x) =∫ 1

0∇ f ((1− t)x+ ty) · (y− x)dt = ∇ f ((1− c)x+ cy) · (y− x)

for some c ∈ (0,1). This is a special case of Taylor’s formula when k = 0.

Appendix J

The Whitney Decomposition of Open Sets in Rn

J.1 Decomposition of Open Sets

An arbitrary open set in Rn can be decomposed as a union of disjoint cubes whose

lengths are proportional to their distance from the boundary of the open set. See, for

instance, Figure J.1 when the open set is the unit disk in R2. For a given cube Q in

Rn, we denote by ℓ(Q) its length.

Proposition. Let Ω be an open nonempty proper subset of Rn. Then there exists a

family of closed dyadic cubes Q j j (called the Whitney cubes of Ω ) such that

(a)⋃

j Q j =Ω and the Q j’s have disjoint interiors.

(b)√

nℓ(Q j)≤ dist (Q j,Ωc)≤ 4

√nℓ(Q j). Thus 10

√nQ j meets Ω c.

(c) If the boundaries of two cubes Q j and Qk touch, then

1

4≤ ℓ(Q j)

ℓ(Qk)≤ 4 .

(d) For a given Q j there exist at most 12n−4n cubes Qk that touch it.

(e) Let 0 < ε < 1/4. If Q∗j has the same center as Q j and ℓ(Q∗j) = (1+ε)ℓ(Q j) then

χΩ =∑j

χQ∗j ≤ 12n−4n +1 .

Proof. Let Dk be the collection of all dyadic cubes of the form

(x1, . . . ,xn) ∈ Rn : m j2−k ≤ x j < (m j +1)2−k ,

where m j ∈ Z. Observe that each cube in Dk gives rise to 2n cubes in Dk+1 by

bisecting each side.

Write the set Ω as the union of the sets

Ωk = x ∈Ω : 2√

n2−k < dist(x,Ω c)≤ 4√

n2−k



609

610 J The Whitney Decomposition of Open Sets in Rn

over all k ∈ Z. Let F ′ be the set of all cubes Q in Dk for some k ∈ Z such that

Q∩Ωk = /0. We show that the collection F ′ satisfies property (b). Let Q ∈F ′ and

pick x ∈Ωk ∩Q for some k ∈ Z. Observe that

√n2−k ≤ dist(x,Ω c)−

√nℓ(Q)≤ dist(Q,Ω c)≤ dist(x,Ω c)≤ 4

√n2−k ,

which proves (b).

Next we observe that ⋃

Q∈F ′Q =Ω .

Indeed, every Q in F ′ is contained in Ω (since it has positive distance from its

complement) and every x ∈Ω lies in some Ωk and in some dyadic cube in Dk.

Fig. J.1 The Whitney decom-

position of the unit disk.

The problem is that the cubes in the collection F ′ may not be disjoint. We have

to refine the collection F ′ by eliminating those cubes that are contained in some

other cubes in the collection. Recall that two dyadic cubes have disjoint interiors or

else one contains the other. For every cube Q in F ′ we can therefore consider the

unique maximal cube Qmax in F ′ that contains it. Two different such maximal cubes

must have disjoint interiors by maximality. Now set F = Qmax : Q ∈F ′.The collection of cubes Q j j =F clearly satisfies (a) and (b), and we now turn

our attention to the proof of (c). Observe that if Q j and Qk in F touch then

√nℓ(Q j)≤ dist(Q j,Ω

c)≤ dist(Q j,Qk)+dist(Qk,Ωc)≤ 0+4

√nℓ(Qk) ,

which proves (c). To prove (d), note that any cube Q in Dk is touched by exactly

3n−1 other cubes in Dk. But each cube Q in Dk can contain at most 4n cubes of F

of length at least one-quarter of the length of Q. This fact combined with (c) yields

(d). To prove (e), notice that each Q∗j is contained in Ω by part (b). If x ∈ Ω , then

J.2 Partition of Unity adapted to Whitney cubes 611

x ∈ Qk0for some k0. If Q j does not touch Qk0

, then Q∗j does not touch Qk0as well.

Consequently, the given x ∈Ω may lie only in Q∗k for these cubes Qk that touch Qk0

and there are 12n−4n +1 such cubes including Qk0.

J.2 Partition of Unity adapted to Whitney cubes

Let us fix an ε such that 0 < ε < 1/4. For each cube Q we denote by Q∗ the cube

with the same center as Q and side length (1+ε)ℓ(Q), where ℓ(Q) is the side length

of Q. Let us fix a nonnegative smooth function φ that is equal to 1 on the unit cube

Q0 = [−1/2,1/2]n and equal to zero outside Q∗0.

Let Qkk be the family of Whitney cubes ofΩ . We denote by ck the center of Qk

and by ℓk its side length. Since dist (Qk,Ωc)≥√nℓk, and dist ((Q∗k)

c,Qk)≤ ε√

nℓk,

it follows that

dist (Q∗k ,Ωc)≥ (1− ε)

√nℓk > 0,

hence Q∗k is contained in Ω . Since the union of Qk is Ω , then the union of Q∗k is

also Ω .

For each k we define

φk(x) = φ(x− ck

ℓk

)

and we set

Φ(x) =∑k

φk(x) .

We notice that Φ is smooth and that Φ ≥ 1 on Ω . Then we define

ϕk(x) =φk(x)

Φ(x).

Obviously ϕk are supported in Q∗k , the union of Q∗k is Ω , and we have

∑k

ϕk = χΩ .

We would like to control |∂αϕk(x)| in terms of ℓk. To achieve this we use the

Leibniz rule of differentiation. For a fixed k we have that

∂α

∂xαϕk(x) = ∑

β≤α

(α1

β1

)· · ·

(αn

βn

)(∂ β

∂xβ

1

Φ(x)

)∂α−β

∂xα−βφk(x)

and obviously∣∣∂α−βφk(x)

∣∣≤ ℓ−|α |+|β |k . A simple inductive argument shows that

∣∣∣ ∂β

∂xβ

1

Φ(x)

∣∣∣≤C ∑0≤q j≤|β |

q1+···+qn=|β |

∣∣∂ q11 · · ·∂

qnn Φ(x)

∣∣Φ(x)2

,

612 J The Whitney Decomposition of Open Sets in Rn

for some constant C, and since for a given x ∈ Ω , Φ(x) is the sum of at most 12n

functions with nonzero values, it follows that

∣∣∂ q11 · · ·∂ qn

n Φ(x)∣∣≤Cn ℓ

−(q1+···+qn)k

when x ∈ Q∗k and thus∣∣∣ ∂

β

∂xβ

1

Φ(x)

∣∣∣≤C′n,β ℓ−|β |k

for x ∈ Q∗k . We conclude that for every multiindex α there is a constant Cα ,n such

that ∣∣∣ ∂α

∂xαϕk(x)

∣∣∣≤Cα ,n ℓ−|α | .

More on Whitney decompositions can be found in the article of Whitney [373]

and the books of Stein [338], Krantz and Parks [204].

Glossary

A B A is a subset of B (also denoted by A⊆ B)

A B A is a proper subset of B

A⊃ B B is a proper subset of A

Ac the complement of a set A

χE the characteristic function of the set E

d f the distribution function of a function f

f ∗ the decreasing rearrangement of a function f

fn ↑ f fn increases monotonically to a function f

Z the set of all integers

Z+ the set of all positive integers 1,2,3, . . .Zn the n-fold product of the integers

R the set of real numbers

R+ the set of positive real numbers

Rn the Euclidean n-space

Q the set of rationals

Qn the set of n-tuples with rational coordinates

C the set of complex numbers

Cn the n-fold product of complex numbers

T the unit circle identified with the interval [0,1]

Tn the n-dimensional torus [0,1]n

|x|√|x1|2 + · · ·+ |xn|2 when x = (x1, . . . ,xn) ∈ Rn



613

614 Glossary

Sn−1 the unit sphere x ∈ Rn : |x|= 1e j the vector (0, . . . ,0,1,0, . . . ,0) with 1 in the jth entry and 0 elsewhere

log t the logarithm with base e of t > 0

loga t the logarithm with base a of t > 0 (1 = a > 0)

log+ t max(0, log t) for t > 0

[t] the integer part of the real number t

x · y the quantity ∑nj=1 x jy j when x = (x1, . . . ,xn) and y = (y1, . . . ,yn)

B(x,R) the ball of radius R centered at x in Rn

ωn−1 the surface area of the unit sphere Sn−1

vn the volume of the unit ball x ∈ Rn : |x|< 1|A| the Lebesgue measure of the set A⊆ Rn

dx Lebesgue measure

AvgB f the average 1|B|

∫B f (x)dx of f over the set B

⟨f ,g

⟩the real inner product

∫Rn f (x)g(x)dx

⟨f |g

⟩the complex inner product

∫Rn f (x)g(x)dx

⟨u, f

⟩the action of a distribution u on a function f

p′ the number p/(p−1), whenever 0 < p = 1 < ∞

1′ the number ∞

∞′ the number 1

f = O(g) means | f (x)| ≤M|g(x)| for some M for x near x0

f = o(g) means | f (x)| |g(x)|−1 → 0 as x→ x0

At the transpose of the matrix A

A∗ the conjugate transpose of a complex matrix A

A−1 the inverse of the matrix A

O(n) the space of real matrices satisfying A−1 = At

‖T‖X→Y the norm of the (bounded) operator T : X → Y

A≈ B means that there exists a c > 0 such that c−1 ≤ BA≤ c

|α| indicates the size |α1|+ · · ·+ |αn| of a multi-index α = (α1, . . . ,αn)

∂mj f the mth partial derivative of f (x1, . . . ,xn) with respect to x j

∂α f ∂α11 · · ·∂αn

n f

Glossary 615

C k the space of functions f with ∂α f continuous for all |α| ≤ k

C0 the space of continuous functions with compact support

C00 the space of continuous functions that vanish at infinity

C ∞0 the space of smooth functions with compact support

D the space of smooth functions with compact support

S the space of Schwartz functions

S0 the space of Schwartz functions ϕ with the property∫

Rn xγϕ(x)dx = 0

for all multi-indices γ .

C ∞ the space of smooth functions⋃∞

k=1 C k

D ′(Rn) the space of distributions on Rn

S ′(Rn) the space of tempered distributions on Rn

E ′(Rn) the space of distributions with compact support on Rn

P the set of all complex-valued polynomials of n real variables

S ′(Rn)/P the space of tempered distributions on Rn modulo polynomials

ℓ(Q) the side length of a cube Q in Rn

∂Q the boundary of a cube Q in Rn

Lp(X ,µ) the Lebesgue space over the measure space (X ,µ)

Lp(Rn) the space Lp(Rn, | · |)Lp,q(X ,µ) the Lorentz space over the measure space (X ,µ)

Lploc(R

n) the space of functions that lie in Lp(K) for any compact set K in Rn

|dµ | the total variation of a finite Borel measure µ on Rn

M (Rn) the space of all finite Borel measures on Rn

Mp(Rn) the space of Lp Fourier multipliers, 1≤ p≤ ∞

M p,q(Rn) the space of translation-invariant operators that map Lp(Rn) to Lq(Rn)

‖µ‖M∫

Rn |dµ | the norm of a finite Borel measure µ on Rn

M the centered Hardy–Littlewood maximal operator with respect to balls

M the uncentered Hardy–Littlewood maximal operator with respect to balls

Mc the centered Hardy–Littlewood maximal operator with respect to cubes

Mc the uncentered Hardy–Littlewood maximal operator with respect to

cubes

616 Glossary

Mµ the centered maximal operator with respect to a measure µ

Mµ the uncentered maximal operator with respect to a measure µ

Ms the strong maximal operator

Md the dyadic maximal operator

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Index

C k functions on the torus, 175

σ -finite, 2

absolutely summable Fourier series, 200

adjoint of an operator, 150

admissible growth, 40

almost everywhere convergence, 285

almost orthogonality, 459, 471

analytic family of operators, 40

Aoki–Rolewicz theorem, 76

A1 condition, 502

Ap condition, 503

necessity of, 543

approximate identity, 26

asymptotics

of Bessel function, 580

asymptotics of gamma function, 567

atom

in a measure space, 56

bad function, 356

Banach–Alaoglou theorem, 601

Banach-valued extension of a linear operator,

391

Banach-valued extension of an operator, 396

Banach-valued measurable function, 392

Banach-valued singular integral, 402

band limited function, 490

Bernoulli polynomials, 299

Bernstein’s inequality, 133

Bernstein’s theorem, 191

Bessel function, 168, 573

asymptotics, 580

large arguments, 579

small arguments, 578

beta function, 564

beta integral identity, 145

Boas and Bochner inequality, 391

Bochner integral, 394

Bochner–Riesz means, 244

Bochner–Riesz operator, 244

Borel measures, 2

bounded variation, 199

BV , functions of bounded variation, 199

Calderon–Zygmund operator, 532

Calderon–Zygmund decomposition, 356

Calderon–Zygmund decomposition on Lq, 373

Calderon–Zygmund operator

maximal, 533

cancellation condition

for a kernel, 375

Carleson operator, 286

Cauchy sequence in measure, 9

centered Hardy–Littlewood maximal function,

86

centered maximal function with respect to

cubes, 99

Cesaro means, 184, 216

characteristic constant of a pair of weights, 524

Characteristic constant of a weight

Ap, 503

characteristic constant of a weight

A1, 503

characterization

of A1 weights, 519

of A∞ weights, 527, 531

Chebyshev’s inequality, 6

circular Dirichlet kernel, 178

circular means, 250

circular partial sum, 184

class (Ap,Ap), 524

class of A1 weights, 502

class of Ap weights, 503



633

634 Index

closed

under translations, 146

closed graph theorem, 602

commutes with translations, 146

compactly supported distribution, 120

complete orthonormal system, 185

completeness

of Lorentz spaces, 54

completeness of Lp, p < 1, 13

conditional expectation, 464

cone multiplier, 158

conjugate function, 245

conjugate harmonic, 318

conjugate harmonic functions, 253

conjugate Poisson kernel, 253, 318, 331

continuously differentiable function

of order N, 104

continuously differentiable functions

on the torus, 175

convergence

in C ∞0 , 119

in S , 106, 119

in Lp, 7

in C ∞, 119

in measure, 6

in weak Lp, 7

convex sequence, 193

convolution, 20, 125

Cotlar’s inequality, 364

countably simple, 56

countably simple function, 2

countably subadditive operator, 47

covering lemma, 88

critical Bochner–Riesz index, 252

critical point, 162

Darboux’s equation, 476

de la Vallee Poussin kernel, 182

decay of Fourier coefficients, 193

decomposition of open sets, 609

decreasing rearrangement, 48

of a simple function, 49

degree of a trigonometric polynomial, 177

deLeeuw’s theorem, 157

derivative

of a distribution, 123

of a function (partial), 104

differentiation

of approximate identities, 96

theory, 93

dilation

L1 dilation, 90

of a function, 109

of a tempered distribution, 124

Dini condition, 384

Dini’s theorem, 212

Dirac mass, 121

directional Hilbert transform, 339

Dirichlet kernel, 178

Dirichlet problem, 92

on the sphere, 144

distribution, 120

homogeneous, 132, 136

of lattice points, 305

supported at a point, 134

tempered, 120

with compact support, 120, 129

distribution function, 3

of a simple function, 4

distributional derivative, 123

divergence

of Bochner–Riesz means at the critical

index, 261

of the Fourier series of a continuous

function, 210

of the Fourier series of an L1 function, 255

doubling condition

on a measure, 98

doubling measure, 98, 505

doubly truncated kernel, 363

doubly truncated singular integrals, 363

duals of Lorentz spaces, 57

duBois Reymond’s theorem, 210

duplication formula

for the gamma function, 570

dyadic cube, 464

dyadic decomposition, 428

dyadic interval, 464

dyadic martingale difference operator, 465

dyadic martingale square function, 469

dyadic maximal function, 103

dyadic spherical maximal function, 461

eigenvalues

of the Fourier transform, 116

equidistributed, 48

equidistributed sequence, 302

essential infimum, 502

Euler’s constant, 569

Euler’s limit formula for the gamma function,

568

infinite product form, 569

expectation

conditional, 464

extrapolation, 47

extrapolation theorem, 548, 553

factorization of Ap weights, 546

Fejer kernel, 27, 180

Index 635

Fejer means, 184

Fejer’s theorem, 205

finite Borel measure, 2

finitely simple function, 2

Fourier coefficient, 175

of a measure, 175

Fourier inversion, 112, 185

on L1, 117

on L2, 114

Fourier multiplier, 155

Fourier series, 176

Fourier transform

of a radial function, 577

of a Schwartz function, 108

of surface measure, 577

on L1, 113

on L2, 113

on Lp, 1 < p < 2, 114

properties of, 109

Frechet space, 106

frame

tight, 496

Fresnel integral, 145

fundamental solution, 135

fundamental theorem of algebra, 143

g-function, 461, 477

gamma function, 563

asymptotics, 567

duplication formula, 570

meromorphic extension, 566

generalized Cauchy-Riemann equations, 332

good function, 356, 372

good lambda inequality, 533

consequences of, 539

gradient, 104

gradient condition

for a kernel, 359

Green’s identity, 135

Holder condition, 384

Holder’s inequality, 3, 11

for weak spaces, 16

Hormander’s condition, 359

Hormander–Mihlin multiplier theorem, 446

Haar function, 465

Haar measure, 18

Haar wavelet, 483

Hadamard’s three lines lemma, 39

Hahn–Banach theorem, 601

Hardy’s inequalities, 31

Hardy–Littlewood maximal function

centered, 86

uncentered, 87

harmonic distribution, 135

harmonic function, 92

Hausdorff–Young, 191

Hausdorff–Young inequality, 114

heat equation, 295

heat kernel, 296

Heisenberg group, 19

Hilbert transform, 314

maximal, 322, 339

maximal directional, 339

truncated, 314

Hirschman’s lemma, 41

homogeneous distribution, 132, 136

homogeneous Lipschitz functions, 195

homogeneous maximal singular integrals, 333

homogeneous singular integrals, 333

improved integrability

of Ap weights, 518

inductive limit topology, 120

infinitely differentiable function, 104

inhomogeneous Lipschitz functions, 196

inner product

complex, 150

real, 150

inner regular measure, 98

interpolation

Banach-valued Marcinkiewicz theorem, 400

Banach-valued Riesz–Thorin theorem, 399

for analytic families of operators, 41

Marcinkiewicz theorem, 33

off-diagonal Marcinkiewicz theorem, 61

Riesz–Thorin theorem, 37

Stein’s theorem, 41

with change of measure, 78

inverse Fourier transform, 112

isoperimetric inequality, 17

Jensen’s inequality, 12, 525

Jones’s factorization theorem, 546

Khintchine’s inequalities, 586

for weak type spaces, 589

Kolmogorov’s inequality, 100

Kolmogorov’s theorem, 255

Kronecker’s lemma, 255

lacunary Carleson–Hunt theorem, 454

lacunary sequence, 227

Laplace’s equation, 328

Laplacian, 135

lattice points, 305

Lebesgue constants, 182

Lebesgue differentiation theorem, 95, 101

636 Index

left Haar measure, 18

left maximal function, 102

Leibniz’s rule, 105, 131

linear operator, 33

Liouville’s theorem, 143

Lipschitz condition, 384

for a kernel, 359

Lipschitz space

homogeneous, 195

inhomogeneous, 196

Littlewood–Paley operator, 420

Littlewood–Paley theorem, 421

localization principle, 213

locally compact topological group, 18

locally finite measure, 98

locally integrable functions, 11

logconvexity of weak L1, 78

Lorentz spaces, 52

M. Riesz’s theorem, 247

majorant

radial decreasing, 92

Marcinkiewicz and Zygmund theorem, 386

Marcinkiewicz function, 415

Marcinkiewicz interpolation theorem, 33

Marcinkiewicz multiplier theorem

on Rn, 441

on R, 439

maximal Calderon–Zygmund operator, 533

maximal function

centered with respect to cubes, 99

dyadic, 103

dyadic spherical, 461

Hardy–Littlewood centered, 86

Hardy–Littlewood uncentered, 87

left, 102

right, 102

strong, 101

uncentered with respect to cubes, 99

with respect to a general measure, 98

maximal Hilbert transform, 322

maximal singular integral, 334

doubly truncated, 363

maximal singular integrals with even kernels,

347

maximal truncated singular integral, 363

measurable set, 2

method of rotations, 339

metrizability

of Lorentz space Lp,q, 74

of weak Lp, 14

Mihlin-Hormander multiplier theorem, 446

minimally supported frequency wavelet, 486

minimax lemma, 556, 603

Minkowski’s inequality, 3, 12, 21

integral form, 13

Minkowski’s integral inequality, 13

Muckenhoupt characteristic constant, 503

Muckenhoupt characteristic constant of a

weight, 503

multi-index, 104

multiplier, 155

on the torus, 272

multiplier theorems, 437

necessity of Ap condition, 543

nonatomic measure space, 56

nonsmooth Littlewood–Paley theorem, 427,

428

normability

p-normability of weak Lp for p < 1, 16

of Lorentz space Lp,q, 74

of Lorentz spaces, 74

of weak Lp for p > 1, 14

normability of weak L1 (lack of), 15

off-diagonal Marcinkiewicz interpolation

theorem, 61

open mapping theorem, 602

operator

commuting with translations, 146

of strong type (p,q), 33

of weak type (p,q), 33

orthonormal set, 185, 484

orthonormal system

complete, 185

oscillation of a function, 94

oscillatory integral, 161

overshoot, 221, 224

Parseval’s relation, 112, 186

partial derivative, 104

partition of unity, 611

phase, 161

Plancherel’s identity, 112, 186

pointwise convergence of Fourier series, 204

Poisson kernel, 26, 92, 95, 183, 317

Poisson kernel for the sphere, 144

Poisson representation formula

of Bessel functions, 573

Poisson summation formula, 187

polarization, 186

positive operator, 397

power weights, 506

principal value integral, 314

principle of localization, 213

quasi-linear operator, 33, 61

quasi-subadditive operator, 45

Index 637

Rademacher functions, 585

radial decreasing majorant, 92

radial function, 90

real analytic function, 262

reflection

of a function, 109


reflection formula

for the gamma function, 570

regulated function, 275

restricted weak type, 77

restricted weak type (1,1), 81

restricted weak type (p,q) with constant C, 61

reverse Holder condition

of order q, 523

reverse Holder property of Ap weights, 514

RHq(µ), 523

Riemann zeta function, 301

Riemann’s principle of localization, 213

Riemann–Lebesgue lemma, 114, 193, 215

Riesz product, 232

Riesz projection, 245

Riesz transform, 325

Riesz–Thorin interpolation theorem, 37

right maximal function, 102

right Haar measure, 18

rotations method, 339

rough singular integrals, 455

Rubio de Francia’s extrapolation theorem, 548

sampling theorem, 491

Schwartz function, 105

Schwartz seminorm, 105

self-adjoint operator, 151

self-transpose operator, 151

Sidon set, 235

simple function, 2

sine integral function, 218

singular integrals with even kernels, 341

size condition

for a kernel, 358, 374

smooth bump, 120

smooth function, 104, 119

smooth function with compact support, 104,

119

smoothly truncated maximal singular integral,

374

smoothness condition

for a kernel, 359, 375

space

L∞, 2

Lp, 2

Lp,∞, 5

Lp,q, 52

M 1,1(Rn), 153

M 2,2(Rn), 154

M ∞,∞(Rn), 154

M p,q(Rn), 151

Mp(Rn), 155

C N , 104

C ∞, 104

C ∞0 , 104

spectrum of the Fourier transform, 116

spherical maximal function, 475

spherical average, 476

spherical coordinates, 591

spherical Dirichlet kernel, 178

spherical partial sum, 184

square Dirichlet kernel, 178

square function, 421, 458

dyadic martingale, 469

square function of Littlewood–Paley, 421

square partial sum, 184

standard kernel, 532

Stein’s interpolation theorem, 41

stereographic projection, 594

Stirling’s formula, 567

stopping-time, 97

stopping-time argument, 356

strong maximal function, 101

strong type (p,q), 33

sublinear operator, 33, 61

summation by parts, 599

support of a distribution, 125

surface area of the unit sphere Sn−1, 565

Taylor formula, 607

tempered distribution, 120

test function, 119

tight frame, 496

tiling of Rn, 428

topological group, 18

torus, 174

total order of differentiation, 104

transference of maximal multipliers, 281

transference of multipliers, 275

translation

of a function, 109


translation operator, 271

translation-invariant operator, 146

transpose of an operator, 150

trigonometric monomial, 177

trigonometric polynomial, 177

truncated Hilbert transform, 314

truncated maximal singular integral, 374

truncated singular integral, 334, 363

638 Index

uncentered Hardy–Littlewood maximal

function, 87

uncentered maximal function with respect to a

general measure, 98

uncentered maximal function with respect to

cubes, 99

uncertainty principle, 118

uniform boundedness principle, 602

unitary matrix, 387

Vandermonde determinant, 165

variation of a function, 199

vector-valued

extrapolation theorem, 554

Hardy–Littlewood maximal inequality, 412,

557

inequalities, 555

vector-valued extension of a linear operator,

390

vector-valued inequalities, 408, 409, 414

vector-valued Littlewood–Paley theorem, 426

vector-valued singular integral, 401

volume of the unit ball in Rn, 565

wave equation, 475

wavelet, 482

of minimally supported frequency, 486

wavelet transform, 496

weak Lp, 5

weak type (1,1), 88

weak type (p,q), 33

Weierstrass approximation theorem, 32

for trigonometric polynomials, 183

Weierstrass’s theorem, 229

weight, 499

of class A1, 502

of class A∞, 525

of class Ap, 503

weighted estimates

for singular integral operators, 540

Weyl’s theorem, 302

Whitney decomposition, 609

Young’s covering lemma, 98

Young’s inequality, 22, 400

for weak type spaces, 23, 73

loukas grafakos classical fourier analysis

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